SATHEE: Chapter 05 Arithmetic Progressions

Chapter 05 Arithmetic Progressions

Multiple Choice Questions (MCQs)

1 In an AP, if $d = - 4, n = 7$ and $a_{n} = 4$ , then $a$ is equal to

(a) 6 (b) 7 (c) 20 (d) 28

Show Answer

Solution

(d) In an AP, $a_{n} = a + (n - 1) d$

$\begin{aligned} 4 = a + (7 - 1) (- 4) & [by given conditions] \\ \Rightarrow & 4 = a + (7 - 1) (- 4) \\ \Rightarrow & 4 = a + 6 (- 4) \\ \Rightarrow & 4 + 24 = a \\ ∴ & a = 28 \end{aligned}$

2 In an AP, if

a = 3.5, d = 0

and

n = 101

, then

a_{n}

will be

(a) 0 (b) 3.5 (c) 103.5 (d) 104.5

Show Answer

Solution

(b) For an AP, $a_{n} = a + (n - 1) d = 3.5 + (101 - 1) \times 0$ [by given conditions]

$∴ = 3.5$

3 The list of numbers

- 10, - 6, - 2, 2, \dots

is (a) an AP with

d = - 16

(b) an AP with

d = 4

d = - 4

(d) not an AP

Show Answer

Solution

(b) The given numbers are $- 10, - 6, - 2, 2, \dots$

$\begin{aligned} Here, a_{1} = - 10, a_{2} = - 6, a_{3} = - 2 and a_{4} = 2, . . \\ Since, \begin{aligned} a_{2} - a_{1} & = - 6 - (- 10) \\ = - 6 + 10 = 4 \\ a_{3} - a_{2} & = - 2 - (- 6) \\ = - 2 + 6 = 4 \\ a_{4} - a_{3} & = 2 - (- 2) \\ = 2 + 2 = 4 \end{aligned} \end{aligned}$

Each successive term of given list has same difference i.e., 4 So, the given list forms an AP with common difference, $d = 4$ .

4 The 11th term of an AP

- 5, \frac{- 5}{2}, 0, \frac{5}{2}, \dots

(a) -20 (b) 20 (c) -30 (d) 30

Show Answer

Solution

(b) Given AP, $- 5, - \frac{5}{2}, 0, \frac{5}{2}$

Here, $a = - 5, d = \frac{- 5}{2} + 5 = \frac{5}{2}$

$\begin{aligned} ∴ a_{11} & = a + (11 - 1) d \\ = - 5 + (10) \times \frac{5}{2} \\ = - 5 + 25 = 20 \end{aligned}$

5 The first four terms of an AP whose first term is -2 and the common difference is -2 are (a)

- 2, 0, 2, 4

(b)

- 2, 4, - 8, 16

(c)

- 2, - 4, - 6, - 8

(d)

- 2, - 4, - 8, - 16

Show Answer

Solution

Given, that first term, $a = - 2$ and common difference, $d = - 2$ , then we have an AP as follows

$\begin{matrix} - 2, - 2 - 2, - 2 + 2 (- 2), - 2 + 3 (- 2) \\ = - 2, - 4, - 6, - 8 \end{matrix}$

6 The 21st term of an AP whose first two terms are -3 and 4, is (a) 17 (b) 137 (c) 143 (d) -143

Show Answer

Solution

(b) Given, first two terms of an AP are $a = - 3$ and $a + d = 4$ .

$\Rightarrow - 3 + d = 4$

Common difference, $d = 7$

$\begin{aligned} ∴ a_{21} & = a + (21 - 1) d \\ = - 3 + (20) 7 \\ = - 3 + 140 = 137 \end{aligned}$

7 If the 2nd term of an AP is 13 and 5th term is 25, what is its 7th term?

(a) 30 (b) 33 (c) 37 (d) 38

Show Answer

Solution

(b) Given, $a_{2} = 13$ and $a_{5} = 25$

$\begin{aligned} \Rightarrow & a + (2 - 1) d & = 13 & [∵ a_{n} = a + (n - 1) d] \\ and & a + (5 - 1) d = 25 \\ \Rightarrow & a + d = 13 & \dots (i) \\ a + 4 d = 25 & \dots (i i) \end{aligned}$

On subtracting Eq. (i) from Eq. (ii), we get

$3 d = 25 - 13 = 12 \Rightarrow d = 4$

From Eq. (i), $a = 13 - 4 = 9$

$∴ a_{7} = a + (7 - 1) d = 9 + 6 \times 4 = 33$

8 Which term of an AP :

21, 42, 63, 84, \dots

is 210 ?

(a) 9 th (b) 10th (c) 11 th (d) 12 th

Show Answer

Solution

(b) Let $n$ th term of the given AP be 210 .

Here, first term,

$a = 21$

and common difference, $d = 42 - 21 = 21$ and $a_{n} = 210$

$\begin{matrix} ∵ & a_{n} = a + (n - 1) d \\ \Rightarrow & 210 = 21 + (n - 1) 21 \\ \Rightarrow & 210 = 21 + 21 n - 21 \\ \Rightarrow & 210 = 21 n \Rightarrow n = 10 \end{matrix}$

Hence, the 10th term of an AP is 210.

9 If the common difference of an AP is 5 , then what is

a_{18} - a_{13}

(a) 5 (b) 20 (c) 25 (d) 30

Show Answer

Solution

$Now, \begin{aligned} a_{18} - a_{13} & = a + (18 - 1) d - [a + (13 - 1) d] [∵ a_{n} = a + (n - 1) d] \\ = a + 17 \times 5 - a - 12 \times 5 \\ = 85 - 60 = 25 \end{aligned}$

10 What is the common difference of an AP in which

a_{18} - a_{14} = 32

(a) 8 (b) -8 (c) -4 (d) 4

Show Answer

Solution

(a) Given,

$a_{18} - a_{14} = 32$

$\Rightarrow a + (18 - 1) d - [a + (14 - 1) d] = 32$

$[∵ a_{n} = a + (n - 1) d]$

$\Rightarrow$

$a + 17 d - a - 13 d = 32$

$\begin{aligned} \Rightarrow & 4 d & = 32 \\ ∴ & d & = 8 \end{aligned}$

which is the required common difference of an AP.

11 Two APs have the same common difference. The first term of one of these is -1 and that of the other is -8 . The difference between their 4th terms is

(a) -1 (b) -8 (c) 7 (d) -9

Show Answer

Solution

By condition, $d_{1} = d_{2} = d$

Let the first term of first AP $(a_{1}) = - 1$

and the first term of second AP $(a_{2}) = - 8$

We know that, the $n$ th term of an AP, $T_{n} = a + (n - 1) d$

$∴$ 4th term of first AP, $T_{4} = a_{1} + (4 - 1) d = - 1 + 3 d$

and 4th term of second AP, $T_{4}^{'} = a_{2} + (4 - 1) d = - 8 + 3 d$

Now, the difference between their 4 th terms is i.e.,

$\begin{aligned} | T_{4} - T_{4}^{'} | & = (- 1 + 3 d) - (- 8 + 3 d) \\ = - 1 + 3 d + 8 - 3 d = 7 \end{aligned}$

Hence, the required difference is 7 .

12 If 7 times the 7 th term of an AP is equal to 11 times its 11 th term, then its 18 th term will be

(a) 7 (b) 11 (c) 18 (d) 0

Show Answer

Solution

(d) According to the question,

$\begin{aligned} 7 a_{7} = 11 a_{11} \\ \Rightarrow 7 [a + (7 - 1) d] = 11 [a + (11 - 1) d] [∵ a_{n} = a + (n - 1) d] \\ \Rightarrow 7 (a + 6 d) = 11 (a + 10 d) \\ \Rightarrow 7 a + 42 d = 11 a + 110 d \\ \Rightarrow 4 a + 68 d = 0 \\ \Rightarrow 2 (2 a + 34 d) = 0 \\ \Rightarrow 2 a + 34 d = 0 [∵ 2 \neq 0] \\ \Rightarrow a + 17 d = 0 \dots (i) \\ ∴ 18th term of an AP, a_{18} = a + (18 - 1) d \\ = a + 17 d = 0 \end{aligned}$

13 The 4th term from the end of an AP

- 11, - 8, - 5, \dots, 49

(a) 37 (b) 40 (c) 43 (d) 58

Show Answer

Solution

(b) We know that, the $n$ th term of an AP from the end is

$\begin{matrix} (i) & a_{n} = l - (n - 1) d \end{matrix}$

Here, $l =$ Last term and $l = 49$

Common difference,

$d = - 8 - (- 11)$

$= - 8 + 11 = 3$

From Eq. (i), $a_{4} = 49 - (4 - 1) 3 = 49 - 9 = 40$

14 The famous mathematician associated with finding the sum of the first 100 natural numbers is

(a) Pythagoras (b) Newton (c) Gauss (d) Euclid

Show Answer

Solution

(c) Gauss is the famous mathematician associated with finding the sum of the first 100 natural numbers i.e., $1, 2, 3, \dots, 100$ .

15 If the first term of an AP is -5 and the common difference is 2, then the sum of the first 6 terms is

(a) 0 (b) 5 (c) 6 (d) 15

Show Answer

Solution

(a) Given,

$\begin{matrix} a = - 5 and d = 2 \\ S_{6} = \frac{6}{2} [2 a + (6 - 1) d] \\ = 3 [2 (- 5) + 5 (2)] \\ = 3 (- 10 + 10) = 0 \end{matrix}$

16 The sum of first 16 terms of the AP

10, 6, 2, \dots

(a) -320 (b) 320 (c) -352 (d) -400

Show Answer

Solution

(a) Given, AP is $10, 6, 2, \dots$

Here, first term $a = 10$ , common difference, $d = - 4$

$\begin{aligned} ∴ S_{16} & = \frac{16}{2} [2 a + (16 - 1) d] ∵ S_{n} = \frac{n}{2} 2 a + (n - 1) d \\ = 8 [2 \times 10 + 15 (- 4)] \\ = 8 (20 - 60) = 8 (- 40) = - 320 \end{aligned}$

17 In an

A P

, if

a = 1, a_{n} = 20

and

S_{n} = 399

, then

n

is equal to

(a) 19 (b) 21 (c) 38 (d) 42

Show Answer

Solution

$\begin{aligned} 399 = \frac{n}{2} [2 \times 1 + (n - 1) d] \\ 798 = 2 n + n (n - 1) d \\ and a_{n} = 20 \\ \Rightarrow a + (n - 1) d = 20 [∵ a_{n} = a + (n - 1) d] \\ \Rightarrow 1 + (n - 1) d = 20 \Rightarrow (n - 1) d = 19 \\ Using Eq. (ii) in Eq. (i), we get \\ \begin{aligned} 798 & = 2 n + 19 n \\ 798 & = 21 n \\ n & = \frac{798}{21} = 38 \end{aligned} \end{aligned}$

$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

18 The sum of first five multiples of 3 is

(a) 45 (b) 55 (c) 65 (d) 75

Show Answer

Solution

(a) The first five multiples of 3 are 3, 6, 9, 12 and 15.

Here, first term, $a = 3$ , common difference, $d = 6 - 3 = 3$ and number of terms, $n = 5$

$\begin{aligned} ∴ S_{5} & = \frac{5}{2} [2 a + (5 - 1) d] ∵ S_{n} = \frac{n}{2} 2 a + (n - 1) d \\ = \frac{5}{2} [2 \times 3 + 4 \times 3] \\ = \frac{5}{2} (6 + 12) = 5 \times 9 = 45 \end{aligned}$

Very Short Answer Type Questions

1 Which of the following form of an AP ? Justify your answer.

(i) $- 1, - 1, - 1, - 1, \dots$ (ii) $0, 2, 0, 2, \dots$

(iii) $1, 1, 2, 2, 3, 3, \dots$ (iv) $11, 22, 33, \dots$

(v) $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots$ (vi) $2, 2^{2}, 2^{3}, 2^{4}$

(vii) $\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \dots$

Show Answer

Solution

(i) Here, $t_{1} = - 1, t_{2} = - 1, t_{3} = - 1$ and $t_{4} = - 1$

$\begin{aligned} t_{2} - t_{1} = - 1 + 1 = 0 \\ t_{3} - t_{2} = - 1 + 1 = 0 \\ t_{4} - t_{3} = - 1 + 1 = 0 \end{aligned}$

Clearly, the difference of successive terms is same, therefore given list of numbers form an AP.

(ii) Here, $t_{1} = 0, t_{2} = 2, t_{3} = 0$ and $t_{4} = 2$

$\begin{aligned} t_{2} - t_{1} = 2 - 0 = 2 \\ t_{3} - t_{2} = 0 - 2 = - 2 \\ t_{4} - t_{3} = 2 - 0 = 2 \end{aligned}$

Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

(iii) Here, $t_{1} = 1, t_{2} = 1, t_{3} = 2$ and $t_{4} = 2$

$\begin{aligned} t_{2} - t_{1} = 1 - 1 = 0 \\ t_{3} - t_{2} = 2 - 1 = 1 \\ t_{4} - t_{2} = 2 - 2 = 0 \end{aligned}$

Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

(iv) Here, $t_{1} = 11, t_{2} = 22$ and $t_{3} = 33$

$\begin{aligned} t_{2} - t_{1} = 22 - 11 = 11 \\ t_{3} - t_{2} = 33 - 22 = 11 \\ t_{4} - t_{3} = 33 - 22 = 11 \end{aligned}$

Clearly, the difference of successive terms is same, therefore given list of numbers form an AP.

(v) $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$

Here,

$\begin{aligned} t_{1} & = \frac{1}{2}, t_{2} = \frac{1}{3} and t_{3} = \frac{1}{4} \\ t_{2} - t_{1} & = \frac{1}{3} - \frac{1}{2} = \frac{2 - 3}{6} = - \frac{1}{6} \\ t_{3} - t_{2} & = \frac{1}{4} - \frac{1}{3} = \frac{3 - 4}{12} = - \frac{1}{12} \end{aligned}$

Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP. (vi) $2, 2^{2}, 2^{3}, 2^{4}, \dots$ i.e., $2, 4, 8, 16, \dots$

Here, $t_{1} = 2, t_{2} = 4, t_{3} = 8$ and $t_{4} = 16$

$\begin{aligned} t_{2} - t_{1} = 4 - 2 = 2 \\ t_{3} - t_{2} = 8 - 4 = 4 \\ t_{4} - t_{3} = 16 - 8 = 8 \end{aligned}$

Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

(vii) $\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \dots$ i.e., $\sqrt{3}, 2 \sqrt{3}, 3 \sqrt{3}, 4 \sqrt{3}, \dots$

Here, $t_{1} = \sqrt{3}, t_{2} = 2 \sqrt{3}, t_{3} = 3 \sqrt{3}$ and $t_{4} = 4 \sqrt{3}$

$t_{2} - t_{1} = 2 \sqrt{3} - \sqrt{3} = \sqrt{3}$

$t_{3} - t_{2} = 3 \sqrt{3} - 2 \sqrt{3} = \sqrt{3}$

$t_{4} - t_{3} = 4 \sqrt{3} - 3 \sqrt{3} = \sqrt{3}$

Clearly, the difference of successive terms is same, therefore given list of numbers form an AP.

2 Justify whether it is true to say that

- 1, \frac{- 3}{2}, - 2, \frac{5}{2}, \dots

forms an AP as

a_{2} - a_{1} = a_{3} - a_{2}

Show Answer

Solution

False

Here, $a_{1} = - 1, a_{2} = \frac{- 3}{2}, a_{3} = - 2$ and $a_{4} = \frac{5}{2}$

$\begin{aligned} a_{2} - a_{1} = \frac{- 3}{2} + 1 = - \frac{1}{2} \\ a_{3} - a_{2} = - 2 + \frac{3}{2} = - \frac{1}{2} \\ a_{4} - a_{3} = \frac{5}{2} + 2 = \frac{9}{2} \end{aligned}$

Clearly, the difference of successive terms is not same, all though, $a_{2} - a_{1} = a_{3} - a_{2}$ but $a_{3} - a_{2} \neq a_{4} - a_{3}$ , therefore it does not form an AP.

3 For the AP

- 3, - 7, - 11, \dots

can we find directly

a_{30} - a_{20}

without actually finding

a_{30}

and

a_{20}

? Give reason for your answer.

Show Answer

Solution

True

$∵ n$ th term of an AP, $a_{n} = a + (n - 1) d$

$∴ a_{30} = a + (30 - 1) d = a + 29 d$

and $a_{20} = a + (20 - 1) d = a + 19 d$

Now, $a_{30} - a_{20} = (a + 29 d) - (a + 19 d) = 10 d$

and from given AP common difference, $d = - 7 - (- 3) = - 7 + 3$

$\begin{aligned} = - 4 \\ a_{30} - a_{20} & = 10 (- 4) = - 40 \end{aligned}$

4 Two AP’s have the same common difference. The first term of one AP is 2 and that of the other is 7 . The difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms? Why?

Show Answer

Solution

Let the same common difference of two AP’s is d. Given that, the first term of first AP and second AP are 2 and 7 respectively, then the AP’s are

$and 7, 7 + d, 7 + 2 d, 7 + 3 d, \dots$

$2, 2 + d, 2 + 2 d, 2 + 3 d, \dots$

Now, 10th terms of first and second AP’s are $2 + 9 d$ and $7 + 9 d$ , respectively.

So, their difference is $7 + 9 d - (2 + 9 d) = 5$

Also, 21 st terms of first and second AP’s are $2 + 20 d$ and $7 + 20 d$ , respectively.

So, their difference is $7 + 20 d - (2 + 9 d) = 5$

Also, if the $a_{n}$ and $b_{n}$ are the $n$ th terms of first and second AP.

Then, $. b_{n} - a_{n} = [7 + (n - 1) d)] - [2 + (n - 1) d] = 5$

Hence, the difference between any two corresponding terms of such AP’s is the same as the difference between their first terms.

5 Is 0 a term of the AP

31, 28, 25, \dots

? Justify your answer.

Show Answer

Solution

Let 0 be the $n$ th term of given AP. i.e., $a_{n} = 0$ .

Given that, first term $a = 31$ , common difference, $d = 28 - 31 = - 3$

The $n$ th terms of an AP, is

$\begin{matrix} \Rightarrow & 0 = 31 + (n - 1) (- 3) \\ \Rightarrow & 3 (n - 1) = 31 \\ \Rightarrow & n - 1 = \frac{31}{3} \\ ∴ & n = \frac{31}{3} + 1 = \frac{34}{3} = 11 \frac{1}{3} \end{matrix}$

Since, $n$ should be positive integer. So, 0 is not a term of the given AP.

6 The taxi fare after each $k m$ , when the fare is ₹ 15 for the first $k m$ and ₹ 8 for each additional km, does not form an AP as the total fare (in ₹) after each $k m$ is $15, 8, 8, 8, \dots$ . Is the statement true? Give reasons.

Show Answer

Solution

No, because the total fare (in ₹) after each $k m$ is

$\begin{aligned} 15, (15 + 8), (15 + 2 \times 8), (15 + 3 \times 8), \dots = 15, 23, 31, 39, \dots \\ t_{1} = 15, t_{2} = 23, t_{3} = 31 and t_{4} = 39 \\ Let t_{2} - t_{1} = 23 - 15 = 8 \\ t_{3} - t_{2} = 31 - 23 = 8 \\ t_{4} - t_{3} = 39 - 31 = 8 \end{aligned}$

Since, all the successive terms of the given list have same difference i.e., common difference $= 8$

Hence, the total fare after each $k m$ form an AP.

7 In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers.

(i) The fee charged from a student every month by a school for the whole session, when the monthly fee is ₹ 400 .

(ii) The fee charged every month by a school from classes I to XII, When the monthly fee for class I is ₹ 250 and it increase by ₹ 50 for the next higher class.

(iii) The amount of money in the account of Varun at the end of every year when ₹ 1000 is deposited at simple interest of $10$ per annum.

(iv) The number of bacteria in a certain food item after each second, when they double in every second.

Show Answer

Solution

(i) The fee charged from a student every month by a school for the whole session is $400, 400, 400, 400, \dots$

which form an AP, with common difference $(d) = 400 - 400 = 0$

(ii) The fee charged month by a school from I to XII is $250, (250 + 50), (250 + 2 \times 50), (250 + 3 \times 50), \dots$

i.e., $250, 300, 350, 400, \dots$

which form an AP, with common difference $(d) = 300 - 250 = 50$

(iii) Simple interest $= \frac{Principal \times Rate \times Time}{100}$

$= \frac{1000 \times 10 \times 1}{100} = 100$

So, the amount of money in the account of Varun at the end of every year is

$1000, (1000 + 100 \times 1), (1000 + 100 \times 2), (1000 + 100 \times 3), \dots$

i.e., $1000, 1100, 1200, 1300, \dots$

which form an AP, with common difference $(d) = 1100 - 1000 = 100$

(iv) Let the number of bacteria in a certain food $= x$

Since, they double in every second.

$\begin{matrix} ∴ & x, 2 x, 2 (2 x), 2 (2 \cdot 2 \cdot x), \dots \\ i.e., & x, 2 x, 4 x, 8 x, \dots \\ t_{1} = x, t_{2} = 2 x, t_{3} = 4 x and t_{4} = 8 x \\ t_{2} - t_{1} = 2 x - x = x \\ t_{3} - t_{2} = 4 x - 2 x = 2 x \\ t_{4} - t_{3} = 8 x - 4 x = 4 x \end{matrix}$

Since, the difference between each successive term is not same. So, the list does form an AP.

8 Justify whether it is true to say that the following are the nth terms of an AP.

(i) $2 n - 3$ (ii) $3 n^{2} + 5$ (iii) $1 + n + n^{2}$

Show Answer

Solution

(i) Yes, here $a_{n} = 2 n - 3$

$\begin{matrix} Put n = 1, & a_{1} = 2 (1) - 3 = - 1 \\ Put n = 2, & a_{2} = 2 (2) - 3 = 1 \\ Put n = 3, & a_{3} = 2 (3) - 3 = 3 \\ Put n = 4, & a_{4} = 2 (4) - 3 = 5 \\ List of numbers becomes - 1, 1, 3, \dots \\ Here, \\ \begin{matrix} a_{2} - a_{1} = 1 - (- 1) = 1 + 1 = 2 \\ a_{3} - a_{2} = 3 - 1 = 2 \\ a_{4} - a_{3} = 5 - 3 = 2 \end{matrix} \\ ∴ a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = \dots \end{matrix}$

Hence, $2 n - 3$ is the $n$ th term of an AP.

(ii) No, here $a_{n} = 3 n^{2} + 5$

$\begin{matrix} Put n = 1, & a_{1} = 3 (1)^{2} + 5 = 8 \\ Put n = 2, & a_{2} = 3 (2)^{2} + 5 = 3 (4) + 5 = 17 \\ Put n = 3, & a_{3} = 3 (3)^{2} + 5 = 3 (9) + 5 = 27 + 5 = 32 \end{matrix}$

So, the list of number becomes $8, 17, 32, \dots$

Here,

$\begin{aligned} a_{2} - a_{1} = 17 - 8 = 9 \\ a_{3} - a_{2} = 32 - 17 = 15 \\ a_{2} - a_{1} \neq a_{3} - a_{2} \end{aligned}$

Since, the successive difference of the list is not same. So, it does not form an AP.

(iii) No, here $a_{n} = 1 + n + n^{2}$

Put $n = 1, a_{1} = 1 + 1 + (1)^{2} = 3$

Put $n = 2, a_{2} = 1 + 2 + (2)^{2} = 1 + 2 + 4 = 7$

Put $n = 3, a_{3} = 1 + 3 + (3)^{2} = 1 + 3 + 9 = 13$

So, the list of number becomes $3, 7, 13, \dots$

$\begin{matrix} Here, & a_{2} - a_{1} = 7 - 3 = 4 \\ ∴ & a_{3} - a_{2} = 13 - 7 = 6 \\ ∴ & a_{2} - a_{1} \neq a_{3} - a_{2} \end{matrix}$

Since, the successive difference of the list is not same. So, it does not form an AP.

Short Answer Type Questions

1 Match the AP’s given in column A with suitable common differences given in column B.

	Column A		Column B
$(A_{1})$	$2, - 2, - 6, - 10, \dots$	$(B_{1})$	$\frac{2}{3}$
$(A_{2})$	$a = - 18, n = 10, a_{n} = 0$	$(B_{2})$	-5
$(A_{3})$	$a = 0, a_{10} = 6$	$(B_{3})$	4
$(A_{4})$	$a_{2} = 13, a_{4} = 3$	$(B_{4})$	-4
		$(B_{5})$	2
		$(B_{6})$	$\frac{1}{2}$

Show Answer

Solution

$A_{1} .2, - 2, - 6, - 10$ ,

Here, common difference, $d = - 2 - 2 = - 4$

$A_{2} . ∵$

$\begin{aligned} a_{n} & = a + (n - 1) d \\ 0 & = - 18 + (10 - 1) d \\ 18 & = 9 d \end{aligned}$

$∴$ Common difference, $d = 2$

$A_{3} . ∵$

$a_{10} = 6$

$\Rightarrow a + (10 - 1) d = 6$

$\Rightarrow 0 + 9 d = 6$

$\Rightarrow 9 d = 6 \Rightarrow d = \frac{2}{3}$

$A_{4} \cdot ∵ a_{2} = 13$

$\Rightarrow a + (2 - 1) d = 13$

$[∵ a_{n} = a + (n - 1) d]$

$\Rightarrow$

$\begin{aligned} (i) & a + d & = 13 \\ (ii) & a_{4} & = 3 \Rightarrow a + (4 - 1) d = 3 \end{aligned}$

and

$∴ a + 3 d = 3$

On subtracting Eq. (i) from Eq. (ii), we get

$\begin{aligned} 2 d = - 10 \\ d = - 5 \\ \Rightarrow \\ (A_{1}) \to B_{4}, (A_{2}) \to B_{5}, (A_{3}) \to B_{1} and (A_{4}) \to B_{2} \end{aligned}$

2 Verify that each of the following is an AP and then write its next three terms.

(i) $0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \dots$

(ii) $5, \frac{14}{3}, \frac{13}{3}, 4, \dots$

(iii) $\sqrt{3}, 2 \sqrt{3}, 3 \sqrt{3}, \dots$

(iv) $a + b, (a + 1) + b, (a + 1) + (b + 1), \dots$

(v) a, 2a $+ 1, 3 a + 2, 4 a + 3, \dots$

Show Answer

Solution

(i) Here, $a_{1} = 0, a_{2} = \frac{1}{4}, a_{3} = \frac{1}{2}$ and $a_{4} = \frac{3}{4}$

$\begin{aligned} a_{2} - a_{1} = \frac{1}{4}, a_{3} - a_{2} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}, a_{4} - a_{3} = \frac{3}{4} - \frac{1}{2} = \frac{1}{4} \\ ∴ & a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} \end{aligned}$

Since, the each successive term of the given list has the same difference. So, it forms an AP. The next three terms are, $a_{5} = a_{1} + 4 d$

$\begin{aligned} = 0 + 4 (\frac{1}{4}) = 1, a_{6} = a_{1} + 5 d = 0 + 5 (\frac{1}{4}) = \frac{5}{4} \\ a_{7} & = a + 6 d = 0 + \frac{6}{4} = \frac{3}{2} \end{aligned}$

(ii) Here, $a_{1} = 5, a_{2} = \frac{14}{3}, a_{3} = \frac{13}{3}$ and $a_{4} = 4$

$\begin{aligned} a_{2} - a_{1} = \frac{14}{3} - 5 = \frac{14 - 15}{3} = \frac{- 1}{3}, a_{3} - a_{2} = \frac{13}{3} - \frac{14}{3} = - \frac{1}{3} \\ a_{4} - a_{3} = 4 - \frac{13}{3} = \frac{12 - 13}{3} = \frac{- 1}{3} \\ ∵ a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} \end{aligned}$

Since, the each successive term of the given list has same difference.

It forms an AP.

The next three terms are,

$\begin{aligned} a_{5} = a_{1} + 4 d = 5 + 4 (- \frac{1}{3}) = 5 - \frac{4}{3} = \frac{11}{3} \\ a_{6} = a_{1} + 5 d = 5 + 5 (- \frac{1}{3}) = 5 - \frac{5}{3} = \frac{10}{3} \\ a_{7} = a_{1} + 6 d = 5 + 6 (- \frac{1}{3}) = 5 - 2 = 3 \end{aligned}$

(iii) Here, $a_{1} = \sqrt{3}, a_{2} = 2 \sqrt{3}$ and $a_{3} = 3 \sqrt{3}$

$\begin{aligned} a_{2} - a_{1} = 2 \sqrt{3} - \sqrt{3} = \sqrt{3}, a_{3} - a_{2} = 3 \sqrt{3} - 2 \sqrt{3} = \sqrt{3} \\ ∵ a_{2} - a_{1} = a_{3} - a_{2} = \sqrt{3} = Common difference \end{aligned}$

Since, the each successive term of the given list has same difference.

So, it forms an AP.

The next three terms are,

$\begin{aligned} a_{4} = a_{1} + 3 d = \sqrt{3} + 3 (\sqrt{3}) = 4 \sqrt{3} \\ a_{5} = a_{1} + 4 d = \sqrt{3} + 4 \sqrt{3} = 5 \sqrt{3} \\ a_{6} = a_{1} + 5 d = \sqrt{3} + 5 \sqrt{3} = 6 \sqrt{3} \end{aligned}$

(iv) Here, $a_{1} = a + b, a_{2} = (a + 1) + b, a_{3} = (a + 1) + (b + 1)$

$\begin{aligned} a_{2} - a_{1} & = (a + 1) + b - (a + b) = a + 1 + b - a - b = 1 \\ a_{3} - a_{2} & = (a + 1) + (b + 1) - [(a + 1) + b] \\ = a + 1 + b + 1 - a - 1 - b = 1 \\ ∵ a_{2} - a_{1} & = a_{3} - a_{2} = 1 = Common difference \end{aligned}$

Since, the each successive term of the given list has same difference.

So, it forms an AP.

The next three terms are,

$\begin{aligned} a_{4} = a_{1} + 3 d = a + b + 3 (1) = (a + 2) + (b + 1) \\ a_{5} = a_{1} + 4 d = a + b + 4 (1) = (a + 2) + (b + 2) \\ a_{6} = a_{1} + 5 d = a + b + 5 (1) = (a + 3) + (b + 2) \end{aligned}$

(v) Here, $a_{1} = a, a_{2} = 2 a + 1, a_{3} = 3 a + 2$ and $a_{4} = 4 a + 3$

$\begin{aligned} a_{2} - a_{1} = 2 a + 1 - a = a + 1 \\ a_{3} - a_{2} = 3 a + 2 - 2 a - 1 = a + 1 \\ a_{4} - a_{3} = 4 a + 3 - 3 a - 2 = a + 1 \\ a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = a + 1 = Common difference \end{aligned}$

Since, the each successive term of the given list has same difference.

So, it forms an AP.

The next three terms are,

$\begin{aligned} a_{5} = a + 4 d = a + 4 (a + 1) = 5 a + 4 \\ a_{6} = a + 5 d = a + 5 (a + 1) = 6 a + 5 \\ a_{7} = a + 6 d = a + 6 (a + 1) = 7 a + 6 \end{aligned}$

3 Write the first three terms of the

A P^{'} s

, when a and

d

are as given below

(i) $a = \frac{1}{2}, d = - \frac{1}{6}$

(ii) $a = - 5, d = - 3$

(iii) $a = \sqrt{2}, d = \frac{1}{\sqrt{2}}$

Show Answer

Solution

(i) Given that, first term $(a) = \frac{1}{2}$ and common difference $(d) = - \frac{1}{6}$

$∵ n$ th term of an AP, $T_{n} = a + (n - 1) d$

$∴$ Second term of an AP, $T_{2} = a + d = \frac{1}{2} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$

and third term of an AP, $T_{3} = a + 2 d = \frac{1}{2} - \frac{2}{6} = \frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6} = \frac{1}{6}$

Hence, required three terms are $\frac{1}{2}, \frac{1}{3}, \frac{1}{6}$ .

(ii) Given that, first term $(a) = - 5$ and common difference $(d) = - 3$

$∵ n$ th term of an AP, $T_{n} = a + (n - 1) d$

$∴$ Second term of an AP, $T_{2} = a + d = - 5 - 3 = - 8$

and third term of an AP, $T_{3} = a + 2 d = - 5 + 2 (- 3)$

$= - 5 - 6 = - 11$

Hence, required three terms are $- 5, - 8, - 11$ .

(iii) Given that, first term $(a) = \sqrt{2}$ and common difference $(d) = \frac{1}{\sqrt{2}}$

$∵ n$ th term of an AP, $T_{n} = a + (n - 1) d$

$∴$ Second term of an AP, $T_{2} = a + d = \sqrt{2} + \frac{1}{\sqrt{2}} = \frac{2 + 1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$

and third term of an AP, $T_{3} + a + 2 d = \sqrt{2} + \frac{2}{\sqrt{2}} = \frac{2 + 2}{\sqrt{2}} = \frac{4}{\sqrt{2}}$

Hence, required three terms are $\sqrt{2}, \frac{3}{\sqrt{2}}, \frac{4}{\sqrt{2}}$ .

4 Find

a, b

and

c

such that the following numbers are in

A P, a, 7, b, 23

and

c

Show Answer

Solution

Since $a, 7, b, 23$ and $c$ are in AP.

$∴ 7 - a = b - 7 = 23 - b = c - 23 =$ Common difference

Taking second and third terms, we get

$\begin{matrix} b - 7 = 23 - b \\ \Rightarrow & 2 b = 30 \\ ∴ & b = 15 \end{matrix}$

Taking first and second terms, we get

$\begin{matrix} \Rightarrow & 7 - a = b - 7 \\ \Rightarrow & 7 - a = 15 - 7 \\ ∴ & 7 - a = 8 \\ \Rightarrow & a = - 1 \end{matrix}$

$\Rightarrow 7 - a = 15 - 7 [∵ b = 15]$

Taking third and fourth terms, we get

$\begin{matrix} 23 - b = c - 23 \\ \Rightarrow & 23 - 15 = c - 23 \\ \Rightarrow & 8 = c - 23 \\ \Rightarrow & 8 + 23 = c \Rightarrow c = 31 \\ Hence, & a = - 1, b = 15, c = 31 \end{matrix}$

5 Determine the AP whose fifth term is 19 and the difference of the eighth term from the thirteenth term is 20.

Show Answer

Solution

Let the first term of an AP be a and common differenced.

$\begin{aligned} Given, a_{5} = 19 and a_{13} - a_{8} = 20 [given] \\ ∴ a_{5} = a + (5 - 1) d = 19 and [a + (13 - 1) d] - [a + (8 - 1) d] = 20 [∵ a_{n} = a + (n - 1) d] \\ \Rightarrow a + 4 d = 19 \\ and & a + 12 d - a - 7 d = 20 \Rightarrow 5 d = 20 \\ ∴ d = 4 \end{aligned}$

On putting $d = 4$ in Eq. (i), we get

$\begin{aligned} a + 4 (4) & = 19 \\ a + 16 & = 19 \\ a & = 19 - 16 = 3 \end{aligned}$

So, required AP is $a, a + d, a + 2 d, a + 3 d, \dots$ i.e., $3, 3 + 4, 3 + 2 (4), 3 + 3 (4), \dots$

i.e., $3, 7, 11, 15$ , ..

6 The 26th, 11th and the last terms of an AP are, 0,3 and

- \frac{1}{5}

, respectively. Find the common difference and the number of terms.

Show Answer

Solution

Let the first term, common difference and number of terms of an AP are $a, d$ and $n$ , respectively.

We know that, if last term of an AP is known, then

$\begin{matrix} (i) & l = a + (n - 1) d \end{matrix}$

and $n$ th term of an AP is

$\begin{matrix} (ii) & T_{n} = a + (n - 1) d \end{matrix}$

Given that, 26th term of an AP $= 0$

$\begin{matrix} \Rightarrow & T_{26} = a + (26 - 1) d = 0 & [from Eq.(i)] \\ \Rightarrow & a + 25 d = 0 & \dots (iii) \end{matrix}$

11th term of an $A P = 3$

$\begin{matrix} \Rightarrow & T_{11} = a + (11 - 1) d = 3 & [from Eq.(ii)] \\ \Rightarrow & a + 10 d = 3 & \dots (iv) \end{matrix}$

and last term of an $A P = - 1 / 5$

$\begin{matrix} \Rightarrow & l = a + (n - 1) d & [from Eq.(i)] \\ \Rightarrow & - 1 / 5 = a + (n - 1) d & \dots (v) \end{matrix}$

Now, subtracting Eq. (iv) from Eq. (iii),

$\begin{array}{r} a + 25 d = 0 \\ \underline{} a \underline{+} 10 d = \underline{} 3 \\ - 15 d = - 3 \\ \Rightarrow d = - \frac{1}{5} \end{array}$

Put the value of $d$ in Eq. (iii), we get

$\begin{aligned} a + 25 (- \frac{1}{5}) & = 0 \\ \Rightarrow & a - 5 & = 0 \Rightarrow a = 5 \end{aligned}$

Now, put the value of $a, d$ in Eq. (v), we get

$\begin{matrix} - 1 / 5 = 5 + (n - 1) (- 1 / 5) \\ \Rightarrow & - 1 = 25 - (n - 1) \\ \Rightarrow & - 1 = 25 - n + 1 \\ \Rightarrow & n = 25 + 2 = 27 \end{matrix}$

Hence, the common difference and number of terms are $- 1 / 5$ and 27 , respectively.

7 The sum of the 5 th and the 7 th terms of an AP is 52 and the 10th term is 46. Find the AP.

Show Answer

Solution

Let the first term and common difference of AP are a and $d$ , respectively. According to the question,

$\begin{aligned} a_{5} + a_{7} = 52 and a_{10} = 46 \\ \Rightarrow & a + (5 - 1) d + a + (7 - 1) d = 52 & [∵ a_{n} = a + (n - 1) d] \\ and & a + (10 - 1) d = 46 \\ \Rightarrow & a + 4 d + a + 6 d = 52 \\ and & a + 9 d = 46 \\ \Rightarrow & 2 a + 10 d = 52 \\ and & a + 9 d = 46 \\ \Rightarrow & a + 5 d = 26 \\ a + 9 d & = 46 \end{aligned}$

On subtracting Eq. (i) from Eq. (ii), we get

$4 d = 20 \Rightarrow d = 5$

From Eq. (i),

$a = 26 - 5 (5) = 1$

So, required AP is $a, a + d, a + 2 d, a + 3 d, \dots$ i.e., $1, 1 + 5, 1 + 2 (5), 1 + 3 (5), \dots$ i.e., $1, 6, 11, 16, \dots$

8 Find the 20th term of the AP whose 7 th term is 24 less than the 11 th term, first term being 12 .

Show Answer

Solution

Let the first term, common difference and number of terms of an AP are $a, d$ and $n$ , respectively.

Given that, first term $(a) = 12$ .

Now by condition,

$\begin{matrix} 7 th term (T_{7}) = 11 th term (T_{11}) - 24 \\ [∵ n th term of an AP, T_{n} = a + (n - 1) d] \\ \Rightarrow & a + (7 - 1) d = a + (11 - 1) d - 24 \\ \Rightarrow & a + 6 d = a + 10 d - 24 \\ \Rightarrow & 24 = 4 d \Rightarrow d = 6 \\ ∴ & 20th term of AP, T_{20} = a + (20 - 1) d \\ = 12 + 19 \times 6 = 126 \end{matrix}$

Hence, the required 20th term of an AP is 126 .

9 If the 9th term of an AP is zero, then prove that its 29th term is twice its 19th term.

Show Answer

Solution

Let the first term, common difference and number of terms of an AP are $a, d$ and $n$ respectively.

Given that, $9$ th term of an AP, $T_{9} = 0 [∵ n .$ th term of an AP, $. T_{n} = a + (n - 1) d]$

$\begin{aligned} \Rightarrow & a + (9 - 1) d = 0 \dots (i) \end{aligned}$

$\Rightarrow a + 8 d = 0 \Rightarrow a = - 8 d$

Now, its 19th term, $T_{19} = a + (19 - 1) d$

$= - 8 d + 18 d [from Eq. (i)]$

$= 10 d \dots (ii)$

and its

29th term, $T_{29} = a + (29 - 1) d$

$= - 8 d + 28 d$

$= 20 d = 2 \times (10 d)$

$\Rightarrow$ $T_{29} = 2 \times T_{19}$

[from Eq. (i)]

Hence, its 29th term is twice its 19th term.

Hence proved.

10 Find whether 55 is a term of the AP

7, 10, 13, \dots

or not. If yes, find which term it is.

Show Answer

Solution

Yes, let the first term, common difference and the number of terms of an AP are $a, d$ and $n$ respectively.

Let the $n$ th term of an AP be 55. i.e., $T_{n} = 55$ .

We know that, the $n$ th term of an AP, $T_{n} = a + (n - 1) d$

Given that, first term $(a) = 7$ and common difference $(d) = 10 - 7 = 3$

From Eq. (i),

$55 = 7 + (n - 1) \times 3$

$\Rightarrow 55 = 7 + 3 n - 3 \Rightarrow 55 = 4 + 3 n$

$\Rightarrow 3 n = 51$

$∴ n = 17$

Since, $n$ is a positive integer. So, 55 is a term of the AP.

Now, put the values of $a, d$ and $n$ in Eq. (i),

$\begin{aligned} T_{n} & = 7 + (17 - 1) (3) \\ = 7 + 16 \times 3 = 7 + 48 = 55 \end{aligned}$

Hence, 17th term of an AP is 55 .

11 Determine

k

, so that

k^{2} + 4 k + 8, 2 k^{2} + 3 k + 6

and

3 k^{2} + 4 k + 4

are three consecutive terms of an AP.

Show Answer

Solution

Since, $k^{2} + 4 k + 8, 2 k^{2} + 3 k + 6$ and $3 k^{2} + 4 k + 4$ are consecutive terms of an AP.

$∴ 2 k^{2} + 3 k + 6 - (k^{2} + 4 k + 8) = 3 k^{2} + 4 k + 4 - (2 k^{2} + 3 k + 6) =$ Common difference

$\Rightarrow 2 k^{2} + 3 k + 6 - k^{2} - 4 k - 8 = 3 k^{2} + 4 k + 4 - 2 k^{2} - 3 k - 6$

$\Rightarrow k^{2} - k - 2 = k^{2} + k - 2$

$\Rightarrow - k = k \Rightarrow 2 k = 0 \Rightarrow k = 0$

12 Split 207 into three parts such that these are in AP and the product of the two smaller parts is 4623 .

Show Answer

Solution

Let the three parts of the number 207 are $(a - d)$ , $a$ and $(a + d)$ , which are in AP. Now, by given condition,

$\Rightarrow$ Sum of these parts $= 207$

$\Rightarrow a - d + a + a + d = 207$

$\Rightarrow 3 a = 207$

$a = 69$

Given that, product of the two smaller parts $= 4623$

$\begin{matrix} \Rightarrow & a (a - d) = 4623 \\ \Rightarrow & 69 \cdot (69 - d) = 4623 \\ \Rightarrow & 69 - d = 67 \\ \Rightarrow & d = 69 - 67 = 2 \\ So, & first part = a - d = 69 - 2 = 67, \\ second part = a = 69 \\ and & third part = a + d = 69 + 2 = 71, \end{matrix}$

Hence, required three parts are 67, 69, 71.

13 The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.

Show Answer

Solution

Given that, the angles of a triangle are in AP.

Let $A, B$ and $C$ are angles of a $△ A B C$ .

$\begin{matrix} ∴ & B = \frac{A + C}{2} \\ \Rightarrow & 2 B = A + C & \dots (i) \end{matrix}$

We know that, sum of all interior angles of a $△ A B C = 180^{\circ}$

$A + B + C = 180^{\circ}$

$\begin{aligned} \Rightarrow & 2 B + B = 180^{\circ} \\ \Rightarrow & 3 B = 180^{\circ} \Rightarrow B = 60^{\circ} \end{aligned}$

Let the greatest and least angles are $A$ and $C$ respectively.

$\begin{aligned} A = 2 C & [by condition] \dots (ii) \end{aligned}$

Now, put the values of $B$ and $A$ in Eq. (i), we get

$\begin{aligned} 2 \times 60 = 2 C + C \\ \Rightarrow 120 = 3 C \Rightarrow C = 40^{\circ} \end{aligned}$

Put the value of $C$ in Eq. (ii), we get

$A = 2 \times 40^{\circ} \Rightarrow A = 80^{\circ}$

Hence, the required angles of triangle are $80^{\circ}, 60^{\circ}$ and $40^{\circ}$ .

14 If the

n

th terms of the two AP’s

9, 7, 5, \dots

and

24, 21, 18, \dots

are the same, then find the value of

n

. Also, that term.

Show Answer

Solution

Let the first term, common difference and number of terms of the AP $9, 7, 5, \dots$ are $a_{1}, d_{1}$ and $n_{1}$ , respectively.

$\begin{array}{ll} ∴ Its n th term, & T_{n_{1}}^{'} = a_{1} + (n_{1} - 1) d_{1} \\ \Rightarrow & T_{n_{1}}^{'} = 9 + (n_{1} - 1) (- 2) \\ \Rightarrow & T_{n_{1}}^{'} = 9 - 2 n_{1} + 2 \\ \Rightarrow & T_{n_{1}}^{'} = 11 - 2 n_{1} [∵ n th term of an AP, T_{n} = a + (n - 1) d] \end{array}$

Let the first term, common difference and the number of terms of the AP $24, 21, 18, \dots$ are $a_{2}, d_{2}$ and $n_{2}$ , respectively.

n th terms of the both APs are same, i.e., $T_{n_{1}}^{'} = T_{n_{2}}^{''}$

$11 - 2 n_{1} = 27 - 3 n_{2}$ [from Eqs. (i) and (ii)]

$\Rightarrow n = 16$

$∴ n$ th term of first AP, $T^{'} n_{1} = 11 - 2 n_{1} = 11 - 2$ (16)

$= 11 - 32 = - 21$

and $n$ th term of second AP, $T^{''} n_{2} = 27 - 3 n_{2} = 27 - 3 (16)$

$= 27 - 48 = - 21$

Hence, the value of $n$ is 16 and that term i.e., $n$ th term is -21 .

15 If sum of the 3rd and the 8th terms of an AP is 7 and the sum of the 7 th and 14 th terms is -3 , then find the 10th term.

Show Answer

Solution

Let the first term and common difference of an AP are a and d, respectively. According to the question,

$\begin{aligned} a_{3} + a_{8} = 7 and a_{7} + a_{14} = - 3 \\ \Rightarrow a + (3 - 1) d + a + (8 - 1) d = 7 [∵ a_{n} = a + (n - 1) d] \\ and a + (7 - 1) d + a + (14 - 1) d = - 3 \\ a + 2 d + a + 7 d = 7 \\ and a + 6 d + a + 13 d = - 3 \\ 2 a + 9 d = 7 \\ 2 a + 19 d = - 3 \\ 10 d = - 10 \Rightarrow d = - 1 \\ 2 a + 9 (- 1) = 7 \\ \Rightarrow 2 a - 9 = 7 \\ \Rightarrow 2 a = 16 \Rightarrow a = 8 \\ ∴ a_{10} = a + (10 - 1) d \\ = 8 + 9 (- 1) \\ = 8 - 9 = - 1 \end{aligned}$

16 Find the 12 th term from the end of the AP

- 2, - 4, - 6, \dots, - 100

Show Answer

Solution

Given AP, $- 2, - 4, - 6, \dots, - 100$

Here, first term $(a) = - 2$ , common difference $(d) = - 4 - (- 2) = - 2$ and the last term $(I) = - 100$ .

We know that, the $n$ th term $a_{n}$ of an AP from the end is $a_{n} = l - (n - 1) d$ , where $l$ is the last term and $d$ is the common difference.

$∴$ 12th term from the end,

$\begin{aligned} a_{12} & = - 100 - (12 - 1) (- 2) \\ = - 100 + (11) (2) = - 100 + 22 = - 78 \end{aligned}$

Hence, the 12th term from the end is -78

17 Which term of the AP

53, 48, 43, \dots

is the first negative term?

Show Answer

Solution

Given AP is $53, 48, 43$ ,..

Whose, first term $(a) = 53$ and common difference $(d) = 48 - 53 = - 5$

Let $n$ th term of the AP be the first negative term.

$\begin{aligned} i.e., T_{n} < 0 [∵ n th term of an AP, T_{n} = a + (n - 1) d] \\ (a + (n - 1) d) < 0 \\ \Rightarrow 53 + (n - 1) (- 5) < 0 \\ \Rightarrow 53 - 5 n + 5 < 0 \\ \Rightarrow 58 - 5 n < 0 \Rightarrow 5 n > 58 \\ \Rightarrow n > 11.6 \Rightarrow n = 12 \\ ∴ T_{12} = a + (12 - 1) d = 53 + 11 (- 5) \\ = 53 - 55 = - 2 < .0 \end{aligned}$

18 How many numbers lie between 10 and 300 , which divided by 4 leave a remainder 3 ?

Show Answer

Solution

Here, the first number is 11 , which divided by 4 leave remainder 3 between 10 and 300 . Last term before 300 is 299, which divided by 4 leave remainder 3 .

$\begin{matrix} ∴ 11, 15, 19, 23, \dots, 299 \\ Here, first term (a) = 11, common difference d = 15 - 11 = 4 \\ ∵ n nth term, a_{n} = a + (n - 1) d = l \\ \Rightarrow 299 = 11 + (n - 1) 4 \\ \Rightarrow 299 - 11 = (n - 1) 4 \\ \Rightarrow 4 (n - 1) = 288 \\ \Rightarrow (n - 1) = 72 \\ ∴ n = 73 \end{matrix}$

19 Find the sum of the two middle most terms of an AP

- \frac{4}{3}, - 1, - \frac{2}{3}, \dots, 4 \frac{1}{3}

Show Answer

Solution

Here, first term $(a) = - \frac{4}{3}$ , common difference $(d) = - 1 + \frac{4}{3} = \frac{1}{3}$

and the last term $(l) = 4 \frac{1}{3} = \frac{13}{3}$

$∵ n$ th term of an AP, $l = a_{n} = a + (n - 1) d$

$\begin{matrix} \Rightarrow & \frac{13}{3} = - \frac{4}{3} + (n - 1) \frac{1}{3} \\ \Rightarrow & 13 = - 4 + (n - 1) \\ \Rightarrow & n - 1 = 17 \\ \Rightarrow & n = 18 & [even] \end{matrix}$

So, the two middle most terms are $\frac{n}{2}$ th and $\frac{n}{2} + 1$ th. i.e., $\frac{18}{2}$ th and $\frac{18}{2} + 1$ th terms i.e., 9th and 10 th terms.

$\begin{matrix} ∴ & a_{9} = a + 8 d = - \frac{4}{3} + 8 \frac{1}{3} = \frac{8 - 4}{3} = \frac{4}{3} \\ and & a_{10} = a + 9 d = \frac{- 4}{3} + 9 \frac{1}{3} = \frac{9 - 4}{3} = \frac{5}{3} \end{matrix}$

So, sum of the two middle most terms $= a_{9} + a_{10} = \frac{4}{3} + \frac{5}{3} = \frac{9}{3} = 3$

20 The first term of an AP is -5 and the last term is 45 . If the sum of the terms of the AP is 120 , then find the number of terms and the common difference.

Show Answer

Solution

Let the first term, common difference and the number of terms of an AP are $a, d$ and $n$ respectively.

Given that, first term $(a) = - 5$ and last term $(l) = 45$

Sum of the terms of the $A P = 120 \Rightarrow S_{n} = 120$

We know that, if last term of an AP is known, then sum of $n$ terms of an AP is,

$\begin{matrix} S_{n} = \frac{n}{2} (a + l) \\ \Rightarrow & 120 = \frac{n}{2} (- 5 + 45) \Rightarrow 120 \times 2 = 40 \times n \\ \Rightarrow & n = 3 \times 2 \Rightarrow n = 6 \end{matrix}$

$∴$ Number of terms of an AP is known, then the $n$ th term of an AP is,

“So, the common difference is 10”

$l = a + (n - 1) d$

$\Rightarrow 45 = - 5 + (6 - 1) d$

$50 = 5 d$

$\Rightarrow d = 10$

“d=10”

Hence, number of terms and the common difference of an AP are 6 and 10 respectively.

21 Find the sum

(i) $1 + (- 2) + (- 5) + (- 8) + \dots + (- 236)$

(ii) $4 - \frac{1}{n} + 4 - \frac{2}{n} + 4 - \frac{3}{n} + \dots$ upto $n$ terms.

(iii) $\frac{a - b}{a + b} + \frac{3 a - 2 b}{a + b} + \frac{5 a - 3 b}{a + b} + \dots$ to 11 terms.

Show Answer

Solution

(i) Here, first term $(a) = 1$ and common difference $(d) = (- 2) - 1 = - 3$

$∵$ Sum of $n$ terms of an AP, $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$\begin{matrix} \Rightarrow & S_{n} = \frac{n}{2} [2 \times 1 + (n - 1) \times (- 3)] \\ \Rightarrow & S_{n} = \frac{n}{2} (2 - 3 n + 3) \Rightarrow S_{n} = \frac{n}{2} (5 - 3 n) & \dots (i) \end{matrix}$

We know that, if the last term ( $l$ ) of an AP is known, then

$\begin{aligned} l = a + (n - 1) d \\ \Rightarrow - 236 = 1 + (n - 1) (- 3) [∵ l = - 236, given] \\ \Rightarrow - 237 = - (n - 1) \times 3 \\ \Rightarrow n - 1 = 79 \Rightarrow n = 80 \end{aligned}$

Now, put the value of $n$ in Eq. (i), we get

$\begin{aligned} S_{n} & = \frac{80}{2} [5 - 3 \times 80] = 40 (5 - 240) \\ = 40 \times (- 235) = - 9400 \end{aligned}$

Hence, the required sum is -9400 .

Alternate Method

Given, $a = 1, d = - 3$ and $l = - 236$

$∴$ Sum of $n$ terms of an AP, $S_{n} = \frac{n}{2} [a + l]$

$\begin{matrix} = \frac{80}{2} (1 + (- 236) & [∵ n = 80] \\ = 40 \times (- 235) = - 9400 \end{matrix}$

(ii) Here, first term, $a = 4 - \frac{1}{n}$

Common difference, $d = 4 - \frac{2}{n} - 4 - \frac{1}{n} = \frac{- 2}{n} + \frac{1}{n} = \frac{- 1}{n}$

$∵$ Sum of $n$ terms of an AP, $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$\begin{aligned} \Rightarrow S_{n} & = (\frac{n}{2}) 24 - (\frac{1}{n}) + (n - 1) (\frac{- 1}{n}) \\ = (\frac{n}{2}) 8 - (\frac{2}{n}) - 1 + (\frac{1}{n}) \\ = (\frac{n}{2}) 7 - (\frac{1}{n}) = (\frac{n}{2}) \times \frac{(7 n - 1)}{n} = \frac{(7 n - 1)}{2} \end{aligned}$

(iii) Here, first term $(A) = \frac{a - b}{a + b}$

and common difference, $D = \frac{3 a - 2 b}{a + b} - \frac{a - b}{a + b} = \frac{2 a - b}{a + b}$

$∵$ Sum of $n$ terms of an AP, $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$\begin{aligned} \Rightarrow S_{n} & = (\frac{n}{2}) 2 \frac{(a - b)}{(a + b)} + (n - 1) \frac{(2 a - b)}{(a + b)} \\ = (\frac{n}{2}) \frac{2 a - 2 b + 2 a n - 2 a - b n + b}{a + b} \\ = (\frac{n}{2}) \frac{(2 a n - b n - b)}{(a + b)} \\ ∴ S_{11} & = (\frac{11}{2}) \frac{2 a (11) - b (11) - b}{a + b} \\ = \frac{11 (11 a - 6 b)}{(a + b)} = (\frac{11}{2}) \frac{(22 a - 12 b)}{(a + b)} \end{aligned}$

22 Which term of the AP $- 2, - 7, - 12, \dots$ will be -77 ? Find the sum of this AP upto the term -77 .

Show Answer

Solution

Given, AP $- 2, - 7, - 12, \dots$

Let the $n$ th term of an AP is -77 .

Then, first term $(a) = - 2$ and common difference $(d) = - 7 - (- 2) = - 7 + 2 = - 5$ .

$∵ n$ th term of an AP, $T_{n} = a + (n - 1) d$

$\begin{aligned} \Rightarrow & - 77 & = - 2 + (n - 1) (- 5) \\ \Rightarrow & - 75 & = - (n - 1) \times 5 \\ \Rightarrow & (n - 1) & = 15 \Rightarrow n = 16 . \end{aligned}$

So, the 16th term of the given AP will be -77 .

Now, the sum of $n$ terms of an AP is

$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

So, sum of 16 terms i.e., upto the term -77 .

$i.e., \begin{aligned} S_{16} & = \frac{16}{2} [2 \times (- 2) + (n - 1) (- 5)] \\ = 8 [- 4 + (16 - 1) (- 5)] = 8 (- 4 - 75) \\ = 8 \times - 79 = - 632 \end{aligned}$

Hence, the sum of this AP upto the term -77 is -632 .

23 If

a_{n} = 3 - 4 n

, then show that

a_{1}, a_{2}, a_{3}, \dots

form an AP. Also, find

S_{20}

Show Answer

Solution

Given that, $n$ th term of the series is $a_{n} = 3 - 4 n$

Put $n = 1, a_{1} = 3 - 4 (1) = 3 - 4 = - 1$

Put $n = 2, a_{2} = 3 - 4 (2) = 3 - 8 = - 5$

Put $n = 3, a_{3} = 3 - 4 (3) = 3 - 12 = - 9$

Put $n = 4, a_{4} = 3 - 4 (4) = 3 - 16 = - 13$

So, the series becomes $- 1, - 5, - 9, - 13$ , .

We see that,

i.e.,

$\begin{aligned} a_{2} - a_{1} = - 5 - (- 1) = - 5 + 1 = - 4, \\ a_{3} - a_{2} = - 9 - (- 5) = - 9 + 5 = - 4, \\ a_{4} - a_{3} = - 13 - (- 9) = - 13 + 9 = - 4 \\ a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = \dots = - 4 \end{aligned}$

Since, the each successive term of the series has the same difference. So, it forms an AP.

We know that, sum of $n$ terms of an AP, $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$∴$ Sum of 20 terms of the AP, $S_{20} = \frac{20}{2} [2 (- 1) + (20 - 1) (- 4)]$

$\begin{aligned} = 10 (- 2 + (19) (- 4)) = 10 (- 2 - 76) \\ = 10 \times - 78 = - 780 \end{aligned}$

Hence, the required sum of 20 terms i.e., $S_{20}$ is -780 .

24 In an

A P

, if

S_{n} = n (4 n + 1)

, then find the AP.

Show Answer

Solution

We know that, the $n$ th term of an AP is

$\begin{aligned} a_{n} & = S_{n} - S_{n - 1} \\ a_{n} & = n (4 n + 1) - (n - 1) 4 (n - 1) + 1 \\ a_{n} & = 4 n^{2} + n - (n - 1) (4 n - 3) \\ = 4 n^{2} + n - 4 n^{2} + 3 n + 4 n - 3 = 8 n - 3 \end{aligned} [∵ S_{n} = n (4 n + 1)]$

Put $n = 1, a_{1} = 8 (1) - 3 = 5$

Put $n = 2, a_{2} = 8 (2) - 3 = 16 - 3 = 13$

Put $n = 3, a_{3} = 8 (3) - 3 = 24 - 3 = 21$

Hence, the required AP is $5, 13, 21, \dots$

25 In an

A P

, if

S_{n} = 3 n^{2} + 5 n

and

a_{k} = 164

, then find the value of

k

Show Answer

Solution

$∵ n$ th term of an $A P$ ,

$\begin{aligned} a_{n} = S_{n} - S_{n - 1} \\ = 3 n^{2} + 5 n - 3 (n - 1)^{2} - 5 (n - 1) & [∵ S_{n} = 3 n^{2} + 5 n (given)] \\ = 3 n^{2} + 5 n - 3 n^{2} - 3 + 6 n - 5 n + 5 \\ a_{n} = 6 n + 2 \\ or & a_{k} = 6 k + 2 = 164 & [∵ a_{k} = 164 (given)] \\ \Rightarrow & 6 k = 164 - 2 = 162 \\ ∴ & k = 27 \end{aligned}$

26 If

S_{n}

denotes the sum of first

n

terms of an

A P

, then prove that

$\begin{matrix} (i) & S_{12} = 3 (S_{8} - S_{4}) . \end{matrix}$

Show Answer

Solution

$∵$ Sum of $n$ terms of an AP, $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$\begin{matrix} ∴ & S_{8} = \frac{8}{2} [2 a + (8 - 1) d] = 4 (2 a + 7 d) = 8 a + 28 d \\ and & S_{4} = \frac{4}{2} [2 a + (4 - 1) d] = 2 (2 a + 3 d) = 4 a + 6 d \\ Now, & S_{8} - S_{4} = 8 a + 28 d - 4 a - 6 d = 4 a + 22 d & \dots (i) \\ and & S_{12} = \frac{12}{2} [2 a + (12 - 1) d] = 6 (2 a + 11 d) \\ = 3 (4 a + 22 d) = 3 (S_{8} - S_{4}) \end{matrix}$

27 Find the sum of first 17 terms of an AP whose 4th and 9th terms are -15 and -30 , respectively.

Show Answer

Solution

Let the first term, common difference and the number of terms in an AP are $a, d$ and $n$ , respectively.

We know that, the $n$ th term of an AP, $T_{n} = a + (n - 1) d$

$∴ 4$ th term of an AP, $T_{4} = a + (4 - 1) d = - 15$

[given]

$\Rightarrow a + 3 d = - 15$

and 9th term of an AP, $T_{9} = a + (9 - 1) d = - 30$

$\Rightarrow a + 8 d = - 30$

Now, subtract Eq. (ii) from Eq. (iii), we get

$\begin{aligned} a + 8 d & = - 30 \\ \underline{} a \underline{+} 3 d & = \underset{+}{} - 15 \\ 5 d & = - 15 \\ \Rightarrow d & = - 3 \end{aligned}$

Put the value of $d$ in Eq. (ii), we get

$\begin{aligned} a + 3 (- 3) = - 15 \Rightarrow a - 9 = - 15 \\ \Rightarrow a = - 15 + 9 \Rightarrow a = - 6 \end{aligned}$

$∵$ Sum of first $n$ terms of an AP, $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$∴$ Sum of first 17 terms of an AP, $S_{17} = \frac{17}{2} [2 \times (- 6) + (17 - 1) (- 3)]$

$\begin{aligned} = \frac{17}{2} [- 12 + (16) (- 3)] \\ = \frac{17}{2} (- 12 - 48) = \frac{17}{2} \times (- 60) \\ = 17 \times (- 30) = - 510 \end{aligned}$

Hence, the required sum of first 17 terms of an AP is -510 .

28 If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256 , then find the sum of first 10 terms.

Show Answer

Solution

Let $a$ and $d$ be the first term and common difference, respectively of an AP.

$∵$ Sum of $n$ terms of an AP, $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

Now,

$\Rightarrow \frac{6}{2} [2 a + (6 - 1) d] = 36$

$\Rightarrow 2 a + 5 d = 12$

and $S_{16} = 256$

$\Rightarrow \frac{16}{2} [2 a + (16 - 1) d] = 256$

$\Rightarrow 2 a + 15 d = 32$

On subtracting Eq. (ii) from Eq. (iii), we get

$\begin{aligned} 10 d = 20 \Rightarrow d = 2 \\ 2 a + 5 (2) = 12 \\ \Rightarrow \\ 2 a = 12 - 10 = 2 \\ a = 1 \\ ∴ \\ S_{10} = \frac{10}{2} [2 a + (10 - 1) d] \\ = 5 [2 (1) + 9 (2)] = 5 (2 + 18) \\ = 5 \times 20 = 100 \end{aligned}$

Hence, the required sum of first 10 terms is 100 .

29 Find the sum of all the 11 terms of an AP whose middle most term is 30 .

Show Answer

Solution

Since, the total number of terms $(n) = 11$

[odd]

$∴$ Middle most term $= \frac{(n + 1)}{2}$ th term $= \frac{11 + 1}{2}$ th term $= 6$ th term

Given that,

$\begin{aligned} a_{6} & = 30 \\ (i) & \Rightarrow a + (6 - 1) d & = 30 \end{aligned}$

$\Rightarrow a + 5 d = 30$

$∵$ Sum of $n$ terms of an AP, $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$∴ \begin{aligned} S_{11} & = \frac{11}{2} [2 a + (11 - 1) d] \\ = \frac{11}{2} (2 a + 10 d) = 11 (a + 5 d) \\ = 11 \times 30 = 330 \end{aligned}$

30 Find the sum of last ten terms of the AP

8, 10, 12, \dots, 126

Show Answer

Solution

For finding, the sum of last ten terms, we write the given AP in reverse order. i.e., $126, 124, 122, \dots, 12, 10, 8$

Here, first term $(a) = 126$ , common difference, $(d) = 124 - 126 = - 2$

$\begin{matrix} ∴ S_{10} = \frac{10}{2} [2 a + (10 - 1) d] & [∵ S_{n} = \frac{n}{2} [2 a + (n - 1) d]] \\ 5 {2 (126) + 9 (- 2)} = 5 (252 - 18) \\ = 5 \times 234 = 1170 \end{matrix}$

31 Find the sum of first seven numbers which are multiples of 2 as well as of 9 .

Show Answer

Solution

For finding, the sum of first seven numbers which are multiples of 2 as well as of 9 . Take LCM of 2 and 9 which is 18.

So, the series becomes $18, 36, 54, \dots$

Here, first term $(a) = 18$ , common difference $(d) = 36 - 18 = 18$

$∴ \begin{aligned} S_{7} & = \frac{n}{2} [2 a + (n - 1) d] = \frac{7}{2} [2 (18) + (7 - 1) 18] \\ = \frac{7}{2} [36 + 6 \times 18] = 7 (18 + 3 \times 18) \\ = 7 (18 + 54) = 7 \times 72 = 504 \end{aligned}$

32 How many terms of the AP

- 15, - 13, - 11, \dots

are needed to make the sum -55 ?

Show Answer

Solution

Let $n$ number of terms are needed to make the sum -55 .

Here, first term $(a) = - 15$ , common difference $(d) = - 13 + 15 = 2$

$∵$ Sum of $n$ terms of an AP, $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$\begin{aligned} \Rightarrow - 55 = \frac{n}{2} [2 (- 15) + (n - 1) 2] [∵ S_{n} = - 55 (given)] \\ \Rightarrow - 55 = - 15 n + n (n - 1) \\ \Rightarrow n^{2} - 16 n + 55 = 0 \\ \Rightarrow n^{2} - 11 n - 5 n + 55 = 0 [by factorisation method] \\ \Rightarrow n (n - 11) - 5 (n - 11) = 0 \\ \Rightarrow (n - 11) (n - 5) = 0 \\ ∴ n = 5, 11 \end{aligned}$

Hence, either 5 and 11 terms are needed to make the sum -55 .

33 The sum of the first

n

terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first

2 n

terms of another AP whose first term is -30 and the common difference is 8 . Find

n

Show Answer

Solution

Given that, first term of the first AP $(a) = 8$ and common difference of the first $A P (d) = 20$

Let the number of terms in first AP be $n$ .

$∵$ Sum of first $n$ terms of an AP, $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$∴ S_{n} = \frac{n}{2} [2 \times 8 + (n - 1) 20]$

$\begin{array}{lr} \Rightarrow S_{n} = \frac{n}{2} (16 + 20 n - 20) \\ \Rightarrow S_{n} = \frac{n}{2} (20 n - 4) \\ ∴ S_{n} = n (10 n - 2) & \dots (i) \\ Now, first term of the second AP (a^{'}) = - 30 \\ and common difference of the second AP (d^{'}) = 8 \\ ∴ Sum of first 2 n terms of second AP, S_{2 n} = \frac{2 n}{2} [2 a^{'} + (2 n - 1) d^{'}] \\ \Rightarrow S_{2 n} = n [2 (- 30) + (2 n - 1) (8)] \\ \Rightarrow . S_{2 n} = n [- 60 + 16 n - 8)] \\ \Rightarrow S_{2 n} = n [16 n - 68] \end{array}$

Now, by given condition,

Sum of first $n$ terms of the first AP $=$ Sum of first $2 n$ terms of the second AP

$\begin{matrix} \Rightarrow & S_{n} = S_{2 n} & [from Eqs. (i) and (ii)] \\ \Rightarrow & n (10 n - 2) = n (16 n - 68) \\ \Rightarrow & n [(16 n - 68) - (10 n - 2)] = 0 \\ \Rightarrow & n (16 n - 68 - 10 n + 2) = 0 \\ \Rightarrow & n (6 n - 66) = 0 \\ ∴ & n = 11 & [∴ n \neq 0] \end{matrix}$

Hence, the required value of $n$ is 11 .

34 Kanika was given her pocket money on Jan 1st, 2008. She puts ₹ 1 on day 1 , ₹ 2 on day 2, ₹ 3 on day 3 and continued doing so till the end of the month, from this money into her piggy bank she also spent ₹ 204 of her pocket money, and found that at the end of the month she still had ₹ 100 with her. How much was her pocket money for the month?

Show Answer

Solution

Let her pocket money be ₹ x.

Now, she takes ₹ 1 on day 1 , ₹ 2 on day 2 , ₹ 3 on day 3 and so on till the end of the month, from this money.

i.e., $1 + 2 + 3 + 4 + \dots + 31$ .

which form an AP in which terms are 31 and first term (a) =1, common difference (d) $= 2 - 1 = 1$

$∴$ Sum of first 31 terms $= S_{31}$

$\begin{aligned} Sum of n terms, \\ ∴ S_{31} = \frac{31}{2} [2 \times 1 + (31 - 1) \times 1] \\ = \frac{31}{2} (2 + 30) = \frac{31 \times 32}{2} \\ = 31 \times 16 = 496 \end{aligned}$

So, Kanika takes ₹ 496 till the end of the month from this money.

Also, she spent ₹ 204 of her pocket money and found that at the end of the month she still has ₹ 100 with her.

Now, according to the condition,

$\begin{aligned} (x - 496) - 204 & = 100 \\ x - 700 & = 100 \\ x & = ₹ 800 \end{aligned}$

Hence, ₹ 800 was her pocket money for the month.

35 Yasmeen saves ₹ 32 during the first month, ₹ 36 in the second month and ₹ 40 in the third month. If she continues to save in this manner, in how many months will she save ₹ 2000 ?

Show Answer

Solution

Given that,

Yasmeen, during the first month, saves =₹ 32

During the second month, saves =₹ 36

During the third month, saves =₹ 40

Let Yasmeen saves ₹ 2000 during the $n$ months.

Here, we have arithmetic progression $32, 36, 40, \dots$

First term $(a) = 32$ , common difference $(d) = 36 - 32 = 4$

and she saves total money, i.e., $S_{n} =$ ₹ 2000

We know that, sum of first $n$ terms of an AP is,

$\begin{matrix} S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ \Rightarrow & 2000 = \frac{n}{2} [2 \times 32 + (n - 1) \times 4] \\ \Rightarrow & 2000 = n (32 + 2 n - 2) \\ \Rightarrow & 2000 = n (30 + 2 n) \\ \Rightarrow & 1000 = n (15 + n) \\ \Rightarrow & 1000 = 15 n + n^{2} \\ \Rightarrow & n^{2} + 15 n - 1000 = 0 \\ \Rightarrow & n (n + 40) - 25 (n + 40) = 0 \Rightarrow (n + 40) (n - 25) = 0 \end{matrix}$

$∴ n = 25 [∵ n \neq - 40]$

Hence, in 25 months will she save ₹ 2000.

[since, months cannot be negative]

Long Answer Type Questions

1 The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the sum of the first ten terms of this AP is 235 , find the sum of its first twenty terms.

Show Answer

Solution

Let the first term, common difference and the number of terms of an AP are $a, d$ and $n$ , respectively.

$∵$ Sum of first $n$ terms of an AP, $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$∴$ Sum of first five terms of an AP, $S_{5} = \frac{5}{2} [2 a + (5 - 1) d]$

$\Rightarrow S_{5} = 5 a + 10 d$

and sum of first seven terms of an AP, $S_{7} = \frac{7}{2} [2 a + (7 - 1) d]$

$= \frac{7}{2} [2 a + 6 d] = 7 (a + 3 d)$

$\Rightarrow$

$\begin{matrix} (iii) & S_{7} = 7 a + 21 d \end{matrix}$

Now, by given condition,

$\begin{aligned} S_{5} + S_{7} & = 167 \\ \Rightarrow & 5 a + 10 d + 7 a + 21 d & = 167 \\ (iv) & \Rightarrow & 12 a + 31 d & = 167 \end{aligned}$

Given that, sum of first ten terms of this AP is 235 .

$\begin{matrix} S_{10} = 235 \\ \Rightarrow & \frac{10}{2} [2 a + (10 - 1) d] = 235 \\ \Rightarrow & 5 (2 a + 9 d) = 235 \\ \Rightarrow & 2 a + 9 d = 47 \dots (v) \end{matrix}$

On multiplying Eq. (v) by 6 and then subtracting it into Eq. (vi), we get

$\begin{matrix} 12 a + 54 d = 282 \\ 12 a + 31 d = 167 \\ - - \\ 23 d = 115 \end{matrix}$

$\Rightarrow d = 5$

Now, put the value of $d$ in Eq. (v), we get

$\begin{aligned} 2 a + 9 (5) = 47 \Rightarrow 2 a + 45 = 47 \\ \Rightarrow 2 a = 47 - 45 = 2 \Rightarrow a = 1 \\ Sum of first twenty terms of this AP, S_{20} = \frac{20}{2} [2 a + (20 - 1) d] \\ = 10 [2 \times (1) + 19 \times (5)] = 10 (2 + 95) \\ = 10 \times 97 = 970 \end{aligned}$

Hence, the required sum of its first twenty terms is 970 .

2 Find the

(i) sum of those integers between 1 and 500 which are multiples of 2 as well as of 5 .

(ii) sum of those integers from 1 to 500 which are multiples of 2 as well as of 5 .

(iii) sum of those integers from 1 to 500 which are multiples of 2 or 5 .

Show Answer

Solution

(i) Since, multiples of 2 as well as of $5 = L C M$ of $(2, 5) = 10$

$∴$ Multiples of 2 as well as of 5 between 1 and 500 is $10, 20, 30, \dots, 490$

which form an AP with first term $(a) = 10$ and common difference $(d) = 20 - 10 = 10$

$n th term a_{n} = Last term (l) = 490$

$∴$ Sum of $n$ terms between 1 and 500 ,

$\begin{matrix} S_{n} = \frac{n}{2} [a + l] \dots (i) \\ ∵ & a_{n} = a + (n - 1) d = l \\ \Rightarrow & 10 + (n - 1) 10 = 490 \\ \Rightarrow & (n - 1) 10 = 480 \\ \Rightarrow & n - 1 = 48 \Rightarrow n = 49 \\ From Eq. (i), & S_{49} & = \frac{49}{2} (10 + 490) \\ = \frac{49}{2} \times 500 \\ = 49 \times 250 = 12250 \end{matrix}$

(ii) Same as part (i),

but multiples of 2 as well as of 5 from 1 to 500 is $10, 20, 30, \dots, 500$ .

$∴$ $a = 10, d = 10, a_{n} = l = 500$

$∵$ $a_{n} = a + (n - 1) d = l$

$\Rightarrow 500 = 10 + (n - 1) 10$

$\Rightarrow 490 = (n - 1) 10$

$\Rightarrow n - 1 = 49 \Rightarrow n = 50$

$∵ S_{n} = \frac{n}{2} (a + l)$

$\Rightarrow S_{50} = \frac{50}{2} (10 + 500) = \frac{50}{2} \times 510$

$= 50 \times 255 = 12750$

(iii) Since, multiples of $2$ or $5 =$ Multiple of $2 +$ Multiple of $5 -$ Multiple of $L C M (2, 5)$ i.e., 10 .

$∴$ Multiples of $2$ or $5$ from $1$ to $500$

$=$ “List of multiple of $2$ from $1$ to $500 +$ List of multiple of $5$ from $1$ to $500$ - List of multiple of 10 from $1$ to $500$ ”

$= (2, 4, 6, \dots, 500) + (5, 10, 15, \dots, 500) - (10, 20, \dots, 500)$

All of these list form an AP.

Now, number of terms in first list,

$500 = 2 + (n_{1} - 1) 2 \Rightarrow 498 = (n_{1} - 1) 2$

$\Rightarrow n_{1} - 1 = 249 \Rightarrow n_{1} = 250$

Number of terms in second list,

$500 = 5 + (n_{2} - 1) 5 \Rightarrow 495 = (n_{2} - 1) 5$

$\Rightarrow 99 = (n_{2} - 1) \Rightarrow n_{2} = 100$

and number of terms in third list,

$\begin{aligned} 500 = 10 + (n_{3} - 1) 10 \Rightarrow 490 = (n_{3} - 1) 10 \\ \Rightarrow n_{3} - 1 = 49 \Rightarrow n_{3} = 50 \end{aligned}$

From Eq. (i), Sum of multiples of 2 or 5 from 1 to 500

$\begin{aligned} = Sum of (2, 4, 6, \dots, 500) + Sum of (5, 10, \dots, 500) - Sum of (10, 20, \dots, 500) \\ = \frac{n_{1}}{2} [2 + 500] + \frac{n_{2}}{2} [5 + 500] - \frac{n_{3}}{2} [10 + 500] ∵ S_{n} = \frac{n}{2} (a + l) \\ = \frac{250}{2} \times 502 + \frac{100}{2} \times 505 - \frac{50}{2} \times 510 \\ = 250 \times 251 + 505 \times 50 - 25 \times 510 = 62750 + 25250 - 12750 \\ = 88000 - 12750 = 75250 \end{aligned}$

3 The eighth term of an AP is half its second term and the eleventh term exceeds one-third of its fourth term by 1 . Find the 15th term.

Show Answer

Solution

Let $a$ and $d$ be the first term and common difference of an AP, respectively.

	Now, by given condition,	$a_{8}$
$\Rightarrow$	$a + 7 d$	$= \frac{1}{2} (a + d)$
$\Rightarrow$	$2 a + 14 d$	$= a + d$
$\Rightarrow$	$a + 13 d$	$= 0$

and

$\begin{matrix} \Rightarrow & a + 10 d = \frac{1}{3} [a + 3 d] + 1 \\ \Rightarrow & 3 a + 30 d = a + 3 d + 3 \\ \Rightarrow & 2 a + 27 d = 3 & \dots (ii) \end{matrix}$

From Eqs. (i) and (ii),

$\begin{matrix} 2 (- 13 d) + 27 d = 3 \\ \Rightarrow & - 26 d + 27 d = 3 \\ \Rightarrow & d = 3 \end{matrix}$

From Eq. (i),

$\begin{matrix} a + 13 (3) = 0 \\ \Rightarrow & a = - 39 \\ ∴ & a_{15} = a + 14 d = - 39 + 14 (3) \\ = - 39 + 42 = 3 \end{matrix}$

4 An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three terms is 429 . Find the AP.

Show Answer

Solution

Since, total number of terms $(n) = 37$

$∴$ Middle term $= \frac{37 + 1}{2}$ th term $= 19$ th term

So, the three middle most terms $= 18$ th, 19 th and 20 th.

By given condition,

Sum of the three middle most terms $= 225$

$(a + 17 d) + (a + 18 d) + (a + 19 d) = 22 d (a)$

$\Rightarrow 3 a + 54 d = 225$

$\Rightarrow a + 18 d = 75$

and sum of the last three terms $= 429$

$\Rightarrow a_{35} + a_{36} + a_{37} = 429$

$\Rightarrow (a + 34 d) + (a + 35 d) + (a + 36 d) = 429$

$\Rightarrow 3 a + 105 d = 429$

$\Rightarrow a + 35 d = 143$

On subtracting Eq. (i) from Eq. (ii), we get

$\begin{aligned} 17 d = 68 \\ \Rightarrow d = 4 \\ From Eq. (i), a + 18 (4) = 75 \\ \Rightarrow a = 75 - 72 \\ \Rightarrow a = 3 \end{aligned}$

“Hence, the resulting A.P is $3, 7, 11, \dots .$ "

5 Find the sum of the integers between 100 and 200 that are (i) divisible by 9 . (ii) not divisible by 9 .

Show Answer

Solution

(i) The numbers (integers) between 100 and 200 which is divisible by 9 are 108, 117, 126, … 198.

Let $n$ be the number of terms between 100 and 200 which is divisible by 9 .

Here, $a = 108, d = 117 - 108 = 9$ and $a_{n} = l = 198$

$\begin{aligned} ∵ & a_{n} & = l = a + (n - 1) d \\ \Rightarrow & 198 & = 108 + (n - 1) 9 \\ \Rightarrow & 90 & = (n - 1) 9 \\ \Rightarrow & n - 1 & = 10 \\ \Rightarrow & n & = 11 \end{aligned}$

$∴$ Sum of terms between 100 and 200 which is divisible by 9 ,

$\begin{aligned} S_{n} & = \frac{n}{2} [2 a + (n - 1) d] \\ \Rightarrow S_{11} & = \frac{11}{2} [2 (108) + (11 - 1) 9] = \frac{11}{2} [216 + 90] \\ = \frac{11}{2} \times 306 = 11 \times 153 = 1683 \end{aligned}$

Hence, required sum of the integers between 100 and 200 that are divisible by 9 is 1683 .

(ii) The sum of the integers between 100 and 200 which is not divisible by $9 =$ (sum of total numbers between 100 and 200) -(sum of total numbers between 100 and 200 which is divisible by 9 ).

Here, $a = 101, d = 102 - 101 = 1$ and $a_{n} = l = 199$

$\begin{matrix} ∵ & a_{n} & = l = a + (n - 1) d \\ \Rightarrow & 199 & = 101 + (n - 1) 1 \end{matrix}$

$\Rightarrow (n - 1) = 98 \Rightarrow n = 99$

$∴$ Sum of terms between 100 and 200,

$\begin{aligned} S_{n} & = \frac{n}{2} [2 a + (n - 1) d] \\ \Rightarrow S_{99} & = \frac{99}{2} [2 (101) + (99 - 1) 1] = \frac{99}{2} [202 + 98] \\ = \frac{99}{2} \times 300 = 99 \times 150 = 14850 \end{aligned}$

From Eq. (i), sum of the integers between 100 and 200 which is not divisible by 9

$\begin{matrix} = 14850 - 1683 & from part (i)] \\ = 13167 \end{matrix}$

Hence, the required sum is 13167.

6 The ratio of the 11 th term to the 18 th term of an AP is

2 : 3

. Find the ratio of the 5th term to the 21st term and also the ratio of the sum of the first five terms to the sum of the first 21 terms.

Show Answer

Solution

Let $a$ and $d$ be the first term and common difference of an AP.

Given that,

$\begin{aligned} \Rightarrow a_{11} : a_{18} & = 2 : 3 \\ \frac{a + 10 d}{a + 17 d} & = \frac{2}{3} \\ \Rightarrow 3 a + 30 d & = 2 a + 34 d \\ \Rightarrow a & = 4 d \dots (i) \end{aligned}$

$\begin{aligned} Now, a_{5} = a + 4 d = 4 d + 4 d = 8 d \\ and a_{21} = a + 20 d = 4 d + 20 d = 24 d \\ ∴ a_{5} : a_{21} = 8 d : 24 d = 1 : 3 \\ = \frac{5}{2} [2 (4 d) + 4 d] \\ = \frac{5}{2} (8 d + 4 d) = \frac{5}{2} \times 12 d = 30 d \\ = \frac{21}{2} [2 (4 d) + 20 d] \\ = \frac{21}{2} (28 d) = 294 d \end{aligned}$

So, ratio of the sum of the first five terms to the sum of the first 21 terms

$S_{5} : S_{21} = 30 d : 294 d = 5 : 49$

7 Show that the sum of an AP whose first term is

a

, the second term

b

and

the last term $c$ , is equal to $\frac{(a + c) (b + c - 2 a)}{2 (b - a)}$ .

Show Answer

Solution

Given that, the AP is $a, b, \dots, c$ .

Here, first term $= a$ , common difference $= b - a$

and last term, $l = a_{n} = c$

$\begin{aligned} ∵ a_{n} = l = a + (n - 1) d \\ \Rightarrow (n - 1) = \frac{(c - a)}{(b - a)} \\ \Rightarrow n = \frac{(c - a)}{(b - a)} + 1 \\ \Rightarrow n = \frac{(c - a + b - a)}{(b - a)} = \frac{(c + b - 2 a)}{(b - a)} \\ \Rightarrow Sum of an AP, S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ ∴= \frac{(b + c - 2 a)}{2 (b - a)} 2 a + \frac{(b + c - 2 a)}{(b - a)} - 1 (b - a) \\ = \frac{(b + c - 2 a)}{2 (b - a)} 2 a + \frac{(c - a)}{(b - a)} \cdot (b - a) \dots (i) \\ = \frac{(b + c - 2 a)}{2 (b - a)} (2 a + c - a) \\ = \frac{(b + c - 2 a)}{2 (b - a)} \cdot (a + c) \end{aligned}$

8 Solve the equation

- 4 + (- 1) + 2 + \dots + x = 437

Show Answer

Solution

$∵$ Given equation is, $- 4 - 1 + 2 + \dots + x = 437$

Here, $- 4 - 1 + 2 + \dots + x$ forms an AP with first term $= - 4$ , common difference $= 3$ , $a_{n} = l = x$

$∵ n$ th term of an AP, $a_{n} = l = a + (n - 1) d$

$\Rightarrow x = - 4 + (n - 1) 3$

$\Rightarrow \frac{x + 4}{3} = n - 1 \Rightarrow n = \frac{x + 7}{3}$

$∴$ Sum of an AP, $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$\begin{aligned} S_{n} & = \frac{(x + 7)}{(2 \times 3)} 2 (- 4) + \frac{(x + 4)}{(3)} \cdot 3 \\ = \frac{(x + 7)}{(2 \times 3)} (- 8 + x + 4) = \frac{(x + 7) (x - 4)}{(2 \times 3)} \end{aligned}$

From Eq. (i),

$\begin{aligned} S_{n} & = 437 \\ \Rightarrow \frac{(x + 7) (x - 4)}{2 \times 3} & = 437 \\ \Rightarrow x^{2} + 7 x - 4 x - 28 & = 874 \times 3 \\ \Rightarrow x^{2} + 3 x - 2650 & = 0 \\ x & = \frac{- 3 \pm \sqrt{(3)^{2} - 4 (- 2650)}}{2} \\ = \frac{- 3 \pm \sqrt{9 + 10600}}{2} \\ = \frac{- 3 \pm \sqrt{10609}}{2} = \frac{- 3 \pm 103}{2} = \frac{100}{2}, \frac{- 106}{2} \\ = 50, - 53 \end{aligned}$

[by quadratic formula]

Here, $x$ cannot be negative. i.e., $x \neq - 53$

also, for $x = - 53, n$ will be negative which is not possible

Hence, the required value of $x$ is 50 .

9 Jaspal Singh repays his total loan of ₹ 118000 by paying every month starting with the first installment of ₹ 1000 . If he increases the installment by ₹ 100 every month, what amount will be paid by him in the 30th installment? What amount of loan does he still have to pay after the 30th installment?

Show Answer

Solution

Given that,

Jaspal singh takes total loan =₹ 118000

He repays his total loan by paying every month.

His first installment =₹ 1000

Second installment =1000+100=₹ 1100

Third installment =1100+100=₹ 1200 and so on

Let its 30th installment be $n$ .

Thus, we have $1000, 1100, 1200, \dots$ which form an AP, with

first term $(a) = 1000$ and common difference $(d) = 1100 - 1000 = 100$

$∵ n$ th term of an AP, $T_{n} = a + (n - 1) d$

For 30th installment, $T_{30} = 1000 + (30 - 1) 100$

$= 1000 + 29 \times 100$

$= 1000 + 2900 = 3900$

So, ₹ 3900 will be paid by him in the 30th installment.

He paid total amount upto 30 installments in the following form

$1000 + 1100 + 1200 + \dots + 3900$

First term $(a) = 1000$ and last term $(l) = 3900$

$∴$ Sum of 30 installments, $S_{30} = \frac{30}{2} [a + l]$

$[∵ .$ sum of first $n$ terms of an AP is, $S_{n} = \frac{n}{2} [a + l]$ , where $l =$ last term $]$

$\Rightarrow S_{30} = 15 (1000 + 3900)$

$= 15 \times$ 4900=₹ 73500

$∴$ Total amount he still have to pay after the 30th installment

$= (Amount of loan) - (Sum of 30 installments)$

= 118000-73500= ₹ 44500

Hence, ₹ 44500 still have to pay after the 30th installment.

10 The students of a school decided to beautify the school on the annual day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every

2 m

. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags.

Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance she did cover in completing this job and returning back to collect her books? What is the maximum distance she travelled carrying a flag?

Show Answer

Solution

Given that, the students of a school decided to beautify the school on the annual day by fixing colourful flags on the straight passage of the school.

Given that, the number of flags $= 27$ and distance between each flag $= 2 m$ .

Also, the flags are stored at the position of the middle most flag i.e., 14th flag and Ruchi was given the responsibility of placing the flags. Ruchi kept her books, where the flags were stored i.e., 14th flag and she could carry only one flag at a time.

Let she placed 13 flags into her left position from middle most flag i.e., 14th flag. For placing second flag and return his initial position distance travelled $= 2 + 2 = 4 m$ .

Similarly, for placing third flag and return his initial position, distance travelled $= 4 + 4 = 8 m$

For placing fourth flag and return his initial position, distance travelled $= 6 + 6 = 12 m$ .

For placing fourteenth flag and return his initial position, distance travelled

$= 26 + 26 = 52 m$

Proceed same manner into her right position from middle most flag i.e., 14th flag.

Total distance travelled in that case $= 52 m$

Also, when Ruchi placed the last flag she return his middle most position and collect her books. This distance also included in placed the last flag.

So, these distances form a series.

$\begin{matrix} 4 + 8 + 12 + 16 + \dots + 52 & [for left] \\ and & 4 + 8 + 12 + 16 + \dots + 52 & [for right] \end{matrix}$

$∴$ Total distance covered by Ruchi for placing these flags

$\begin{aligned} = 2 \times (4 + 8 + 12 + \dots + 52) \\ = 2 \times \frac{13}{2} 2 \times 4 + (13 - 1) \times (8 - 4) \frac{∵ Sum of n terms of an AP}{S_{n} = \frac{n}{2} [2 a + (n - 1) d} \\ . = 2 \times \frac{13}{2} (8 + 12 \times 4) ∵ both sides of Ruchi number of flags i.e., n = 13] \\ = 2 \times [13 (4 + 12 \times 2)] = 2 \times 13 (4 + 24) \\ = 2 \times 13 \times 28 = 728 m \end{aligned}$

Hence, the required is $728 m$ in which she did cover in completing this job and returning back to collect her books.

Now, the maximum distance she travelled carrying a flag = Distance travelled by Ruchi during placing the 14th flag in her left position or 27th flag in her right position

$\begin{aligned} = (2 + 2 + 2 + \dots + 13 times) \\ = 2 \times 13 = 26 m \end{aligned}$

Hence, the required maximum distance she travelled carrying a flag is $26 m$ .

Mathematics 10

Chapter 05 Arithmetic Progressions

Multiple Choice Questions (MCQs)

Very Short Answer Type Questions

Short Answer Type Questions

Long Answer Type Questions

Table of Contents

Engineering Preparation

Medical Preparation

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Syllabus

Mentorship

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Mathematics 10

Chapter 05 Arithmetic Progressions

Multiple Choice Questions (MCQs)

Very Short Answer Type Questions

Short Answer Type Questions

Long Answer Type Questions

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Mentorship

NCERT Exercise Video Solution

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