Solution

(d) In an AP, $ \quad\quad\quad $ $ a_n=a+(n-1) d $

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

Solution

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

$\therefore \quad=3.5$

Solution

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

$ \begin{aligned} & \text{ Here, } a_1=-10, a_2=-6, a_3=-2 \text{ and } a_4=2, . . \\ & \text{ Since, } \quad \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, $d=4$.

Solution

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

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

$ \begin{aligned} \therefore \quad a _{11} & =a+(11-1) d \\ & =-5+(10) \times \dfrac{5}{2} \\ & =-5+25=20 \end{aligned} $

Solution

(c) Let the first four terms of an AP are $a, a+d, a+2 d$ and $a+3 d$.

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

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

Solution

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

$\Rightarrow \quad-3+d=4$

Common difference, $d=7$

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

Solution

(b) Given, $a_2=13$ and $a_5=25$

$ \begin{aligned}\Rightarrow & & a+(2-1) d & =13 & {[\because a_n=a+(n-1) d] } \\ \text{ and } & & a+(5-1) d =25 \\ \Rightarrow & & a+d =13 & \ldots (i) \\ & & a+4 d =25 & \ldots (ii) \end{aligned} $

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

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

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

$\therefore \quad a_7=a+(7-1) d=9+6 \times 4=33$

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} \because & 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.

Solution

(c) Given, the common difference of AP i.e., $d=5$

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

Solution

(a) Given,

$ a _{18}-a _{14}=32 $

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

$ [\because a_n=a+(n-1) d] $

$\Rightarrow$

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

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

which is the required common difference of an AP.

Solution

(c) Let the common difference of two APs are $d_1$ and $d_2$, respectively.

By condition, $\quad d_1=d_2=d$

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, $T_n=a+(n-1) d$

$\therefore$ 4th term of first AP, $T_4=a_1+(4-1) d=-1+3 d$

and 4th term of second AP, $T_4{ }^{\prime}=a_2+(4-1) d=-8+3 d$

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

$ \begin{aligned} |T_4-T_4^{\prime}| & =(-1+3 d)-(-8+3 d) \\ & =-1+3 d+8-3 d=7 \end{aligned} $

Hence, the required difference is 7 .

Solution

(d) According to the question,

$ \begin{aligned} & 7 a_7=11 a _{11} \\ & \Rightarrow \quad 7[a+(7-1) d]=11[a+(11-1) d] \quad[\because a_n=a+(n-1) d] \\ & \Rightarrow \quad 7(a+6 d)=11(a+10 d) \\ & \Rightarrow \quad 7 a+42 d=11 a+110 d \\ & \Rightarrow \quad 4 a+68 d=0 \\ & \Rightarrow \quad 2(2 a+34 d)=0 \\ & \Rightarrow \quad 2 a+34 d=0 \quad[\because 2 \neq 0] \\ & \Rightarrow \quad a+17 d=0 \quad \ldots \text{ (i) } \\ & \therefore \quad \text{ 18th term of an AP, } a _{18}=a+(18-1) d \\ & =a+17 d=0 \end{aligned} $

Solution

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

$$ \begin{equation*} a_n=l-(n-1) d \tag{i} \end{equation*} $$

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

Common difference,

$ d=-8-(-11) $

$ =-8+11=3 $

From Eq. (i), $a_4=49-(4-1) 3=49-9=40$

Solution

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

Solution

(a) Given,

$ \begin{matrix} & a =-5 \text{ and } d=2 \\ & S_6 =\dfrac{6}{2}[2 a+(6-1) d] & \\ & =3[2(-5)+5(2)] \\ & =3(-10+10)=0 \end{matrix} $

Solution

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

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

$ \begin{aligned} \therefore \quad S _{16} & =\dfrac{16}{2}[2 a+(16-1) d] \quad \because S_n=\dfrac{n}{2}{2 a+(n-1) d} \\ & =8[2 \times 10+15(-4)] \\ & =8(20-60)=8(-40)=-320 \end{aligned} $

Solution

(c) :

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

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

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} \therefore \quad S_5 & =\dfrac{5}{2}[2 a+(5-1) d] \quad \because S_n=\dfrac{n}{2}{2 a+(n-1) d} \\ & =\dfrac{5}{2}[2 \times 3+4 \times 3] \\ & =\dfrac{5}{2}(6+12)=5 \times 9=45 \end{aligned} $

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) $\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}$

Here,

$$ \begin{aligned} t_1 & =\dfrac{1}{2}, t_2=\dfrac{1}{3} \text{ and } t_3=\dfrac{1}{4} \\ t_2-t_1 & =\dfrac{1}{3}-\dfrac{1}{2}=\dfrac{2-3}{6}=-\dfrac{1}{6} \\ t_3-t_2 & =\dfrac{1}{4}-\dfrac{1}{3}=\dfrac{3-4}{12}=-\dfrac{1}{12} \end{aligned} $$

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

Here, $\quad t_1=2, t_2=4, t_3=8$ 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) $\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \ldots$ i.e., $\sqrt{3}, 2 \sqrt{3}, 3 \sqrt{3}, 4 \sqrt{3}, \ldots$

Here, $\quad t_1=\sqrt{3}, t_2=2 \sqrt{3}, t_3=3 \sqrt{3}$ 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.

Solution

False

Here, $a_1=-1, a_2=\dfrac{-3}{2}, a_3=-2$ and $a_4=\dfrac{5}{2}$

$$ \begin{aligned} & a_2-a_1=\dfrac{-3}{2}+1=-\dfrac{1}{2} \\ & a_3-a_2=-2+\dfrac{3}{2}=-\dfrac{1}{2} \\ & a_4-a_3=\dfrac{5}{2}+2=\dfrac{9}{2} \end{aligned} $$

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.

Solution

True

$\because n$th term of an AP, $a_n=a+(n-1) d$

$\therefore \quad a _{30}=a+(30-1) d=a+29 d$

and $\quad a _{20}=a+(20-1) d=a+19 d$

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

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

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

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

$ \text{ and } \quad 7,7+d, 7+2 d, 7+3 d, \ldots $

$ 2,2+d, 2+2 d, 2+3 d, \ldots $

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.

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 =\dfrac{31}{3} \\ \therefore & & n =\dfrac{31}{3}+1=\dfrac{34}{3}=11 \dfrac{1}{3} \end{matrix} $

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

Solution

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

$$ \begin{aligned} & 15,(15+8),(15+2 \times 8),(15+3 \times 8), \ldots=15,23,31,39, \ldots \\ & t_1=15, t_2=23, t_3=31 \text{ and } t_4=39 \\ & \text{ Let } \quad t_2-t_1=23-15=8 \\ & t_3-t_2=31-23=8 \\ & t_4-t_3=39-31=8 \end{aligned} $$

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

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

Solution

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

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

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

i.e., $\quad 250,300,350,400, \ldots$

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

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

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

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

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

i.e., $\quad 1000,1100,1200,1300, \ldots$

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} \therefore & x, 2 x, 2 (2 x), 2(2 \cdot 2 \cdot x), \ldots \\ \text{ i.e., } & x, 2 x, 4 x, 8 x, \ldots \\ & t_1 =x, t_2=2 x, t_3=4 x \text{ 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.

Solution

(i) Yes, here $a_n=2 n-3$

$ \begin{matrix} \text{ Put } n=1, & a_1=2(1)-3=-1 \\ \text{ Put } n=2, & a_2=2(2)-3=1 \\ \text{ Put } n=3, & a_3=2(3)-3=3 \\ \text{ Put } n=4, & a_4=2(4)-3=5 \\ \text{ List of numbers becomes }-1,1,3, \ldots \\ \text{ Here, } \\ \qquad \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} \\ \therefore \quad a_2-a_1=a_3-a_2=a_4-a_3=\ldots \end{matrix} $

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

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

$ \begin{matrix} \text{ Put } n=1, & a_1=3(1)^{2}+5=8 \\ \text{ Put } n=2, & a_2=3(2)^{2}+5=3(4)+5=17 \\ \text{ Put } n=3, & a_3=3(3)^{2}+5=3(9)+5=27+5=32 \end{matrix} $

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

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, $\quad$ here $a_n=1+n+n^{2}$

Put $n=1, \quad a_1=1+1+(1)^{2}=3$

Put $n=2, \quad a_2=1+2+(2)^{2}=1+2+4=7$

Put $n=3, \quad a_3=1+3+(3)^{2}=1+3+9=13$

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

$ \begin{matrix} \text{ Here, } & a_2-a_1=7-3=4 \\ \therefore & a_3-a_2=13-7=6 \\ \therefore & 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.

Solution

$A_1 .2,-2,-6,-10$,

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

$A_2 . \because$

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

$\therefore$ Common difference, $d=2$

$A_3 . \because$

$$ a _{10}=6 $$

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

$\Rightarrow \quad 0+9 d=6$

$\Rightarrow \quad 9 d=6 \Rightarrow d=\dfrac{2}{3}$

$ A_4 \cdot \because \quad a_2=13 $

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

$[\because a_n=a+(n-1) d]$

$\Rightarrow$

$$ \begin{align*} a+d & =13 \tag{i}\\ a_4 & =3 \Rightarrow a+(4-1) d=3 \tag{ii} \end{align*} $$

and

$\therefore \quad a+3 d=3$

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 \text{ and }(A_4) \to B_2 \end{aligned} $

Solution

(i) Here, $a_1=0, a_2=\dfrac{1}{4}, a_3=\dfrac{1}{2}$ and $a_4=\dfrac{3}{4}$

$ \begin{aligned} & a_2-a_1=\dfrac{1}{4}, a_3-a_2=\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4}, a_4-a_3=\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4} \\ \therefore \quad & 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 \left( \dfrac{1}{4} \right)=1, a_6=a_1+5 d=0+5 \left( \dfrac{1}{4} \right)=\dfrac{5}{4} \\ a_7 & =a+6 d=0+\dfrac{6}{4}=\dfrac{3}{2} \end{aligned} $

(ii) Here, $a_1=5, a_2=\dfrac{14}{3}, a_3=\dfrac{13}{3}$ and $a_4=4$

$$ \begin{aligned} & a_2-a_1=\dfrac{14}{3}-5=\dfrac{14-15}{3}=\dfrac{-1}{3}, a_3-a_2=\dfrac{13}{3}-\dfrac{14}{3}=-\dfrac{1}{3} \\ & a_4-a_3=4-\dfrac{13}{3}=\dfrac{12-13}{3}=\dfrac{-1}{3} \\ & \because \quad a_2-a_1=a_3-a_2=a_4-a_3 \end{aligned} $$

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

It forms an AP.

The next three terms are,

$$ \begin{aligned} & a_5=a_1+4 d=5+4 \left(-\dfrac{1}{3} \right) =5-\dfrac{4}{3}=\dfrac{11}{3} \\ & a_6=a_1+5 d=5+5 \left(-\dfrac{1}{3}\right) =5-\dfrac{5}{3}=\dfrac{10}{3} \\ & a_7=a_1+6 d=5+6 \left(-\dfrac{1}{3}\right) =5-2=3 \end{aligned} $$

(iii) Here, $a_1=\sqrt{3}, a_2=2 \sqrt{3}$ and $a_3=3 \sqrt{3}$

$$ \begin{aligned} & a_2-a_1=2 \sqrt{3}-\sqrt{3}=\sqrt{3}, a_3-a_2=3 \sqrt{3}-2 \sqrt{3}=\sqrt{3} \\ & \because \quad a_2-a_1=a_3-a_2=\sqrt{3}=\text{ Common difference } \end{aligned} $$

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, $a_1=a+b, a_2=(a+1)+b, a_3=(a+1)+(b+1)$

$$ \begin{aligned} a_2-a_1 & =(a+1)+b-(a+b)=a+1+b-a-b=1 \\ a_3-a_2 & =(a+1)+(b+1)-[(a+1)+b] \\ & =a+1+b+1-a-1-b=1 \\ \because \quad a_2-a_1 & =a_3-a_2=1=\text{ Common difference } \end{aligned} $$

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 $a_4=4 a+3$

$$ \begin{aligned} & a_2-a_1=2 a+1-a=a+1 \\ & a_3-a_2=3 a+2-2 a-1=a+1 \\ & a_4-a_3=4 a+3-3 a-2=a+1 \\ & a_2-a_1=a_3-a_2=a_4-a_3=a+1=\text{ Common difference } \end{aligned} $$

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

So, it forms an AP.

The next three terms are,

$$ \begin{aligned} & a_5=a+4 d=a+4(a+1)=5 a+4 \\ & a_6=a+5 d=a+5(a+1)=6 a+5 \\ & a_7=a+6 d=a+6(a+1)=7 a+6 \end{aligned} $$

Solution

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

$\because n$th term of an AP, $T_n=a+(n-1) d$

$\therefore$ Second term of an AP, $T_2=a+d=\dfrac{1}{2}-\dfrac{1}{6}=\dfrac{2}{6}=\dfrac{1}{3}$

and third term of an AP, $T_3=a+2 d=\dfrac{1}{2}-\dfrac{2}{6}=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{3-2}{6}=\dfrac{1}{6}$

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

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

$\because n$th term of an AP, $T_n=a+(n-1) d$

$\therefore$ 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 $(d)=\dfrac{1}{\sqrt{2}}$

$\because n$th term of an AP, $T_n=a+(n-1) d$

$\therefore$ Second term of an AP, $T_2=a+d=\sqrt{2}+\dfrac{1}{\sqrt{2}}=\dfrac{2+1}{\sqrt{2}}=\dfrac{3}{\sqrt{2}}$

and third term of an AP, $T_3+a+2 d=\sqrt{2}+\dfrac{2}{\sqrt{2}}=\dfrac{2+2}{\sqrt{2}}=\dfrac{4}{\sqrt{2}}$

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

Solution

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

$\therefore \quad 7-a=b-7=23-b=c-23=$ Common difference

Taking second and third terms, we get

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

Taking first and second terms, we get

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

$ \Rightarrow \quad 7-a=15-7 \quad[\because 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 \\ \text{ Hence, } & a=-1, b=15, c=31 & \end{matrix} $

Solution

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

$$ \begin{aligned} & \text { Given, } a_5=19 \text { and } a_{13}-a_8=20 \quad \text{[given]} \\ & \therefore \quad a_5=a+(5-1) d=19 \text { and }[a+(13-1) d]-[a+(8-1) d]=20 \quad\left[\because a_n=a+(n-1) d\right] \\ & \Rightarrow \quad a+4 d=19 \\ \text { and } & a+12 d-a-7 d=20 \Rightarrow 5 d=20 \\ & \therefore \quad \quad \quad d=4 \\ & \end{aligned} $$

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 $a, a+d, a+2 d, a+3 d, \ldots$ i.e., $3,3+4,3+2(4), 3+3(4), \ldots$

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

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

$$ \begin{equation*} l=a+(n-1) d \tag{i} \end{equation*} $$

and $n$th term of an AP is

$$ \begin{equation*} T_n=a+(n-1) d \tag{ii} \end{equation*} $$

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

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

11th term of an $AP=3$

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

and last term of an $AP=-1 / 5$

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

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

$$ \begin{aligned} a+25 d =0 \\ \underset{-}{ }a\underset{-}{+}10 d =\underset{-}{ }3 \\ \hline \\ -15 d =-3 \\ \Rightarrow \quad d =-\dfrac{1}{5} \end{aligned} $$

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

$ \begin{aligned} & a+25(-\dfrac{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.

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 \text{ and } a _{10} =46 \\ \Rightarrow & & a+(5-1) d+a+(7-1) d =52 & & [\because a_n=a+(n-1) d]\\ \text{ and } & & a+(10-1) d =46 \\ \Rightarrow & & a+4 d+a+6 d =52 \\ \text{ and } & & a+9 d =46 \\ \Rightarrow & &2 a+10 d =52 \\ \text{ 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 $a, a+d, a+2 d, a+3 d, \ldots$ i.e., $1,1+5,1+2(5), 1+3(5), \ldots \quad$ i.e., $1,6,11,16, \ldots$

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 \text{ th term }(T_7) =11 \text{ th term }(T _{11})-24 \\ & & {[\because n \text{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 \quad d=6 \\ \therefore & \text{ 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 .

Solution

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

Given that, $\quad 9$ th term of an AP, $T_9=0 \quad[\because n.$th term of an AP, $.T_n=a+(n-1) d]$

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

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

Now, its $\quad$ 19th term, $T _{19}=a+(19-1) d$

$ =-8 d+18 d \quad \text{ [from Eq. (i) }] $

$ =10 d \quad\quad \ldots \text{(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.

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),

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

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

$\Rightarrow \quad 3 n=51$

$\therefore \quad n=17$

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 .

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.

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

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

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

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

Solution

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

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

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

$\Rightarrow \quad\quad\quad 3 a=207 $

$ \quad\quad\quad \quad 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 \\ \text{ So, } & & \text{ first part } =a-d=69-2=67, \\ & & \text{ second part } =a=69 \\ \text{ and } & & \text{ third part } = a+d=69+2=71, \end{matrix} $

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

Solution

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

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

$ \begin{matrix}\therefore & B=\dfrac{A+C}{2} \\ \Rightarrow & 2 B=A+C & \ldots \text{(i)} \end{matrix} $

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

$ 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 & &\text{[by condition]} \ldots \text{(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 \quad 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 $40^{\circ}$.

Solution

Let the first term, common difference and number of terms of the AP $9,7,5, \ldots$ are $a_1, d_1$ and $n_1$, respectively.

$\begin{array}{ll}\therefore \text { Its } n \text {th term, } & T_{n_1}^{\prime}=a_1+\left(n_1-1\right) d_1 \\ \Rightarrow & T_{n_1}^{\prime}=9+\left(n_1-1\right)(-2) \\ \Rightarrow & T_{n_1}^{\prime}=9-2 n_1+2 \\ \Rightarrow & T_{n_1}^{\prime}=11-2 n_1 \quad\left[\because n \text {th term of an AP, } T_n=a+(n-1) d\right]\end{array}$

Let the first term, common difference and the number of terms of the AP $24,21,18, \ldots$ are $a_2, d_2$ and $n_2$, respectively.

n th terms of the both APs are same, i.e., $ T _{n_1}^{\prime} = T _{n_2}^{\prime\prime} $

$11-2 n_1=27-3 n_2 \quad$ [from Eqs. (i) and (ii)]

$\Rightarrow \quad n=16$

$\therefore n$th term of first AP, $T^{\prime} n_1=11-2 n_1=11-2$ (16)

$$ =11-32=-21 $$

and $n$th term of second AP, $T^{\prime \prime} n_2=27-3 n_2=27-3(16)$

$$ =27-48=-21 $$

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

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 \text{ and } a_7+a _{14}=-3 \\ & \Rightarrow \quad a+(3-1) d+a+(8-1) d=7 \quad[\because a_n=a+(n-1) d] \\ & \text{ and } \quad a+(7-1) d+a+(14-1) d=-3 \\ & a+2 d+a+7 d=7 \\ & \text{ and } \quad 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 \quad 2 a-9=7 \\ & \Rightarrow \quad 2 a=16 \Rightarrow a=8 \\ & \therefore \quad a _{10}=a+(10-1) d \\ & =8+9(-1) \\ & =8-9=-1 \end{aligned} $

Solution

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

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.

$\therefore$ 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

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} & \text{ i.e., } \quad T_n<0 \quad[\because n \text{th term of an AP, } T_n=a+(n-1) d] \\ & (a+(n-1) d)<0 \\ & \Rightarrow \quad 53+(n-1)(-5)<0 \\ & \Rightarrow \quad 53-5 n+5<0 \\ & \Rightarrow \quad 58-5 n<0 \Rightarrow 5 n>58 \\ & \Rightarrow \quad n>11.6 \Rightarrow n=12 \\ & \therefore \quad T _{12}=a+(12-1) d=53+11(-5) \\ & =53-55=-2<.0 \end{aligned} $

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} \therefore 11,15,19,23, \ldots,299 \\ \text{ Here, first term }(a)=11, \text{ common difference } d=15-11=4 \\ \because n \text{ 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 \\ \therefore n =73 \end{matrix} $

Solution

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

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

$\because n$th term of an AP, $l=a_n=a+(n-1) d$

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

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

$ \begin{matrix} \therefore & a_9=a+8 d=-\dfrac{4}{3}+8 \dfrac{1}{3}=\dfrac{8-4}{3}=\dfrac{4}{3} \\ \text{ and } & a _{10}=a+9 d=\dfrac{-4}{3}+9 \dfrac{1}{3}=\dfrac{9-4}{3}=\dfrac{5}{3} \end{matrix} $

So, sum of the two middle most terms $=a_9+a _{10}=\dfrac{4}{3}+\dfrac{5}{3}=\dfrac{9}{3}=3$

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

$\therefore$ 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.

Solution

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

$\because$ Sum of $n$ terms of an AP, $S_n=\dfrac{n}{2}[2 a+(n-1) d]$

$ \begin{matrix} \Rightarrow & S_n=\dfrac{n}{2}[2 \times 1+(n-1) \times(-3)] \\ \Rightarrow & S_n=\dfrac{n}{2}(2-3 n+3) \Rightarrow S_n=\dfrac{n}{2}(5-3 n) & \ldots \text{(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 \quad-236=1+(n-1)(-3) \quad[\because l=-236 \text{, given }] \\ & \Rightarrow \quad-237=-(n-1) \times 3 \\ & \Rightarrow \quad n-1=79 \Rightarrow n=80 \end{aligned} $

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

$ \begin{aligned} S_n & =\dfrac{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 $l=-236$

$\therefore$ Sum of $n$ terms of an AP, $S_n=\dfrac{n}{2}[a+l]$

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

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

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

$\because$ Sum of $n$ terms of an AP, $S_n=\dfrac{n}{2}[2 a+(n-1) d]$

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

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

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

$\because$ Sum of $n$ terms of an AP, $S_n=\dfrac{n}{2}[2 a+(n-1) d]$

$ \begin{aligned} \Rightarrow \quad S_n & =(\dfrac{n}{2} )2 \dfrac{(a-b)}{(a+b)}+(n-1) \dfrac{(2 a-b)}{(a+b)} \\ & =(\dfrac{n}{2}) \dfrac{2 a-2 b+2 a n-2 a-b n+b}{a+b} \\ & =(\dfrac{n}{2}) \dfrac{(2 a n-b n-b)}{(a+b)} \\ \therefore \quad S _{11} & =(\dfrac{11}{2}) \dfrac{2 a(11)-b(11)-b}{a+b} \\ & =\dfrac{11(11 a-6 b)}{(a+b)}=(\dfrac{11}{2}) \dfrac{(22 a-12 b)}{(a+b)} \end{aligned} $

Solution

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

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

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

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

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

$ \text{ i.e., } \quad \begin{aligned} S _{16} & =\dfrac{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 .

Solution

Given that, $n$th term of the series is $a_n=3-4 n$

Put $n=1, \quad a_1=3-4(1)=3-4=-1$

Put $n=2, \quad a_2=3-4(2)=3-8=-5$

Put $n=3, \quad a_3=3-4(3)=3-12=-9$

Put $n=4, \quad a_4=3-4(4)=3-16=-13$

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

We see that,

i.e.,

$$ \begin{aligned} & a_2-a_1=-5-(-1)=-5+1=-4, \\ & a_3-a_2=-9-(-5)=-9+5=-4, \\ & a_4-a_3=-13-(-9)=-13+9=-4 \\ & a_2-a_1=a_3-a_2=a_4-a_3=\ldots=-4 \end{aligned} $$

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, $S_n=\dfrac{n}{2}[2 a+(n-1) d]$

$\therefore$ Sum of 20 terms of the AP, $S _{20}=\dfrac{20}{2}[2(-1)+(20-1)(-4)]$

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

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

Solution

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

$$ \begin{aligned} a_n & =S_n-S _{n-1} \\ a_n & =n(4 n+1)-(n-1){4(n-1)+1} \\ a_n & =4 n^{2}+n-(n-1)(4 n-3) \\ & =4 n^{2}+n-4 n^{2}+3 n+4 n-3=8 n-3 \end{aligned} \quad[\because S_n=n(4 n+1)] $$

Put $n=1, \quad a_1=8(1)-3=5$

Put $n=2, \quad a_2=8(2)-3=16-3=13$

Put $n=3, \quad a_3=8(3)-3=24-3=21$

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

Solution

$\because n$th term of an $AP$,

$ \begin{array}{rl}&a_n=S_n-S_{n-1} & \\ & =3 n^{2}+5 n-3(n-1)^{2}-5(n-1) & [\because S_n=3 n^{2}+5 n \text{ (given) }] \\ & =3 n^{2}+5 n-3 n^{2}-3+6 n-5 n+5 &\\ & a_n=6 n+2 & \\ \text{ or } & a_k=6 k+2=164 & [\because a_k=164 \text{ (given) }] \\ \Rightarrow & 6 k=164-2=162 & \\ \therefore & k=27 & \end{array} $

Solution

$\because$ Sum of $n$ terms of an AP, $S_n=\dfrac{n}{2}[2 a+(n-1) d]$

$ \begin{matrix} \therefore & S_8 =\dfrac{8}{2}[2 a+(8-1) d]=4(2 a+7 d)=8 a+28 d \\ \text{ and } & S_4 =\dfrac{4}{2}[2 a+(4-1) d]=2(2 a+3 d)=4 a+6 d \\ \text{ Now, } & S_8-S_4 =8 a+28 d-4 a-6 d=4 a+22 d & \ldots \text{(i)} \\ \text{ and } & S _{12} =\dfrac{12}{2}[2 a+(12-1) d]=6(2 a+11 d) \\ & =3(4 a+22 d)=3(S_8-S_4) \end{matrix} $

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, $T_n=a+(n-1) d$

$\therefore \quad 4$ th term of an AP, $T_4=a+(4-1) d=-15$

[given]

$\Rightarrow \quad a+3 d=-15$

and 9th term of an AP, $T_9=a+(9-1) d=-30$

$\Rightarrow \quad a+8 d=-30$

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

$ \begin{aligned} a+8 d & =-30 \\ \underset{-}{ } a \underset{-}{+}3 d & =\underset{+}{ }-15 \\ \hline 5 d & =-15 \\ \Rightarrow \quad 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 \quad a=-15+9 \Rightarrow a=-6 \end{aligned} $

$\because$ Sum of first $n$ terms of an AP, $S_n=\dfrac{n}{2}[2 a+(n-1) d]$

$\therefore$ Sum of first 17 terms of an AP, $S _{17}=\dfrac{17}{2}[2 \times(-6)+(17-1)(-3)]$

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

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

Solution

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

$\because$ Sum of $n$ terms of an AP, $S_n=\dfrac{n}{2}[2 a+(n-1) d]$

Now,

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

$\Rightarrow \quad 2 a+5 d=12$

and $\quad S _{16}=256$

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

$\Rightarrow \quad 2 a+15 d=32$

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 \\ & \therefore \\ & S _{10}=\dfrac{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 .

Solution

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

[odd]

$\therefore \quad$ Middle most term $=\dfrac{(n+1)}{2}$ th term $=\dfrac{11+1}{2}$ th term $=6$ th term

Given that,

$$ \begin{align*} a_6 & =30 \\ \Rightarrow a+(6-1) d & =30 \tag{i} \end{align*} $$

$\Rightarrow \quad a+5 d=30$

$\because \quad$ Sum of $n$ terms of an AP, $S_n=\dfrac{n}{2}[2 a+(n-1) d]$

$ \therefore \quad \begin{aligned} S _{11} & =\dfrac{11}{2}[2 a+(11-1) d] \\ & =\dfrac{11}{2}(2 a+10 d)=11(a+5 d) \\ & =11 \times 30=330 \end{aligned} $

Solution

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

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

$\begin{matrix} \therefore \quad S _{10} =\dfrac{10}{2}[2 a+(10-1) d] & [\because S_n=\dfrac{n}{2}[2 a+(n-1) d]] \\ & 5\{2(126)+9(-2)\}=5(252-18) & \\ & =5 \times 234=1170 & \end{matrix} $

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 $18,36,54, \ldots$

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

$\therefore \quad \begin{aligned} S_7 & =\dfrac{n}{2}[2 a+(n-1) d]=\dfrac{7}{2}[2(18)+(7-1) 18] \\ & =\dfrac{7}{2}[36+6 \times 18]=7(18+3 \times 18) \\ & =7(18+54)=7 \times 72=504\end{aligned}$

Solution

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

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

$\because \quad$ Sum of $n$ terms of an AP, $S_n=\dfrac{n}{2}[2 a+(n-1) d]$

$ \begin{aligned} & \Rightarrow \quad-55=\dfrac{n}{2}[2(-15)+(n-1) 2] \quad[\because S_n=-55 \text{ (given) }] \\ & \Rightarrow \quad-55=-15 n+n(n-1) \\ & \Rightarrow \quad n^{2}-16 n+55=0 \\ & \Rightarrow \quad n^{2}-11 n-5 n+55=0 \quad \text{ [by factorisation method] } \\ & \Rightarrow \quad n(n-11)-5(n-11)=0 \\ & \Rightarrow \quad(n-11)(n-5)=0 \\ & \therefore \quad n=5,11 \end{aligned} $

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

Solution

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

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

$\because$ Sum of first $n$ terms of an AP, $S_n=\dfrac{n}{2}[2 a+(n-1) d]$

$\therefore \quad S_n=\dfrac{n}{2}[2 \times 8+(n-1) 20]$

$ \begin{array}{lr} \Rightarrow S_n=\dfrac{n}{2}(16+20 n-20) \\ \Rightarrow S_n=\dfrac{n}{2}(20 n-4) \\ \therefore S_n=n(10 n-2) & \ldots{\text{(i)}}\\ \text{ Now, first term of the second AP }(a^{\prime})=-30 \\ \text{ and common difference of the second AP }(d^{\prime})=8 \\ \therefore \text{ Sum of first } 2 n \text{ terms of second AP, } S_{2 n}=\dfrac{2 n}{2}[2 a^{\prime}+(2 n-1) d^{\prime}] \\ \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} & \text{ [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 & \\ \therefore & n =11 & [\therefore n \neq 0] \end{matrix} $

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

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., $\quad 1+2+3+4+\ldots+31$.

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

$\therefore$ Sum of first 31 terms $=S _{31}$

$ \begin{aligned} & \text{ Sum of } n \text{ terms, } \\ & \therefore \quad S _{31}=\dfrac{31}{2}[2 \times 1+(31-1) \times 1] \\ & =\dfrac{31}{2}(2+30)=\dfrac{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 & =\text {₹} 800 \end{aligned} $

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

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 $32,36,40, \ldots$

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 =\dfrac{n}{2}[2 a+(n-1) d] \\ \Rightarrow & 2000 =\dfrac{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} $

$ \therefore \quad n=25 \quad[\because n \neq-40] $

Hence, in 25 months will she save ₹ 2000.

[since, months cannot be negative]

Solution

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

$\because$ Sum of first $n$ terms of an AP, $S_n=\dfrac{n}{2}[2 a+(n-1) d]$

$\therefore$ Sum of first five terms of an AP, $S_5=\dfrac{5}{2}[2 a+(5-1) d]$

$ \Rightarrow \quad S_5=5 a+10 d $

and sum of first seven terms of an AP, $S_7=\dfrac{7}{2}[2 a+(7-1) d]$

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

$\Rightarrow$

$$ \begin{equation*} S_7=7 a+21 d \tag{iii} \end{equation*} $$

Now, by given condition,

$$ \begin{align*} & S_5+S_7 & =167 \\ \Rightarrow & 5 a+10 d+7 a+21 d & =167 \\ \Rightarrow & 12 a+31 d & =167 \tag{iv} \end{align*} $$

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

$ \begin{matrix}& S_{10} =235 \\ \Rightarrow & \dfrac{10}{2}[2 a+(10-1) d] =235 \\ \Rightarrow & 5(2 a+9 d) =235 \\ \Rightarrow & 2 a+9 d =47 \ldots \text{(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 \\ -\quad- \\ \hline 23 d=115 \end{matrix} $

$ \Rightarrow \quad 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 \quad 2 a=47-45=2 \Rightarrow a=1 \\ & \text{ Sum of first twenty terms of this AP, } S _{20}=\dfrac{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 .

Solution

(i) Since, multiples of 2 as well as of $5=LCM$ of $(2,5)=10$

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

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

$ n \text{th term } a_n=\text{ Last term }(l)=490 $

$\therefore$ Sum of $n$ terms between 1 and 500 ,

$ \begin{matrix} & S_n =\dfrac{n}{2}[a+l] \ldots \text{(i)} \\ \because & 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 \\ \text{ From Eq. (i), } & S _{49} &=\dfrac{49}{2}(10+490) \\ & =\dfrac{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 $10,20,30, \ldots, 500$.

$\therefore$ $a=10, d=10, a_n=l=500$

$\because$ $a_n=a+(n-1) d=l$

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

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

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

$\because \quad S_n=\dfrac{n}{2}(a+l)$

$\Rightarrow \quad S _{50}=\dfrac{50}{2}(10+500)=\dfrac{50}{2} \times 510$

$=50 \times 255=12750$

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

$\therefore$ 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 $500$”

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

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 $

$\Rightarrow \quad n_1-1=249 \Rightarrow n_1=250$

Number of terms in second list,

$ 500=5+(n_2-1) 5 \Rightarrow 495=(n_2-1) 5 $

$ \Rightarrow \quad 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 \quad 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} & =\text{ Sum of }(2,4,6, \ldots, 500)+\text{ Sum of }(5,10, \ldots, 500)-\text{ Sum of }(10,20, \ldots, 500) \\ & =\dfrac{n_1}{2}[2+500]+\dfrac{n_2}{2}[5+500]-\dfrac{n_3}{2}[10+500] \quad \because S_n=\dfrac{n}{2}(a+l) \\ & =\dfrac{250}{2} \times 502+\dfrac{100}{2} \times 505-\dfrac{50}{2} \times 510 \\ & =250 \times 251+505 \times 50-25 \times 510=62750+25250-12750 \\ & =88000-12750=75250 \end{aligned} $

Solution

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

and

$ \begin{matrix} \Rightarrow & a+10 d=\dfrac{1}{3}[a+3 d]+1 \\ \Rightarrow & 3 a+30 d=a+3 d+3 \\ \Rightarrow & 2 a+27 d=3 & \ldots \text{(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 \\ \therefore & a _{15} =a+14 d=-39+14(3) \\ & =-39+42=3 \end{matrix} $

Solution

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

$\therefore$ Middle term $=\dfrac{37+1}{2}$ 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 $=225$

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

$\Rightarrow \quad 3 a+54 d=225$

$\Rightarrow \quad a+18 d=75$

and sum of the last three terms $=429$

$\Rightarrow \quad a _{35}+a _{36}+a _{37}=429$

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

$\Rightarrow \quad 3 a+105 d=429$

$\Rightarrow \quad a+35 d=143$

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

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

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

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 $a_n=l=198$

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

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

$ \begin{aligned} S_n & =\dfrac{n}{2}[2 a+(n-1) d] \\ \Rightarrow \quad S _{11} & =\dfrac{11}{2}[2(108)+(11-1) 9]=\dfrac{11}{2}[216+90] \\ & =\dfrac{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, $\quad a=101, d=102-101=1$ and $a_n=l=199$

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

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

$\therefore$ Sum of terms between 100 and 200,

$ \begin{aligned} S_n & =\dfrac{n}{2}[2 a+(n-1) d] \\ \Rightarrow \quad S _{99} & =\dfrac{99}{2}[2(101)+(99-1) 1]=\dfrac{99}{2}[202+98] \\ & =\dfrac{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 & \text{ from part }(i)] \\ =13167 & \end{matrix} $

Hence, the required sum is 13167.

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 \\ \dfrac{a+10 d}{a+17 d} & =\dfrac{2}{3} \\ \Rightarrow 3 a+30 d & =2 a+34 d \\ \Rightarrow a & =4 d \ldots \text{(i)} \end{aligned} $

$ \begin{aligned} & \text{ Now, } \quad a_5=a+4 d=4 d+4 d=8 d \\ & \text{ and } \quad a _{21}=a+20 d=4 d+20 d=24 d \\ & \therefore \quad a_5: a _{21}=8 d: 24 d=1: 3 \\ & =\dfrac{5}{2}[2(4 d)+4 d] \\ & =\dfrac{5}{2}(8 d+4 d)=\dfrac{5}{2} \times 12 d=30 d \\ & =\dfrac{21}{2}[2(4 d)+20 d] \\ & =\dfrac{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 $

Solution

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

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

and last term, $l=a_n=c$

$ \begin{aligned} \because a_n =l=a+(n-1) d \\ & & & \\ \Rightarrow (n-1) =\dfrac{(c-a)}{(b-a)} \\ \Rightarrow n =\dfrac{(c-a)}{(b-a)}+1 \\ \Rightarrow n =\dfrac{(c-a+b-a)}{(b-a)}=\dfrac{(c+b-2 a)}{(b-a)} \\ \Rightarrow \text{ Sum of an AP, } S_n =\dfrac{n}{2}[2 a+(n-1) d] \\ \therefore =\dfrac{(b+c-2 a)}{2(b-a)} 2 a+\dfrac{(b+c-2 a)}{(b-a)}-1 \quad(b-a) \\ =\dfrac{(b+c-2 a)}{2(b-a)} 2 a+\dfrac{(c-a)}{(b-a)} \cdot(b-a) \ldots \text{(i)}\\ =\dfrac{(b+c-2 a)}{2(b-a)}(2 a+c-a) \\ =\dfrac{(b+c-2 a)}{2(b-a)} \cdot(a+c) \end{aligned} $

Solution

$\because$ Given equation is, $-4-1+2+\ldots+x=437$

Here, $-4-1+2+\ldots+x$ forms an AP with first term $=-4$, common difference $=3$, $a_n=l=x$

$\because n$th term of an AP, $a_n=l=a+(n-1) d$

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

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

$\therefore \quad$ Sum of an AP, $S_n=\dfrac{n}{2}[2 a+(n-1) d]$

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

From Eq. (i),

$ \begin{aligned} S_n & =437 \\ \Rightarrow \quad \dfrac{(x+7)(x-4)}{2 \times 3} & =437 \\ \Rightarrow \quad x^{2}+7 x-4 x-28 & =874 \times 3 \\ \Rightarrow \quad x^{2}+3 x-2650 & =0 \\ x & =\dfrac{-3 \pm \sqrt{(3)^{2}-4(-2650)}}{2} \\ & =\dfrac{-3 \pm \sqrt{9+10600}}{2} \\ & =\dfrac{-3 \pm \sqrt{10609}}{2}=\dfrac{-3 \pm 103}{2}=\dfrac{100}{2}, \dfrac{-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 .

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 $1000,1100,1200, \ldots$ which form an AP, with

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

$\because n$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+\ldots+3900 $

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

$\therefore$ Sum of 30 installments, $S _{30}=\dfrac{30}{2}[a+l]$

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

$\Rightarrow \quad S _{30}=15(1000+3900)$

$=15 \times $ 4900=₹ 73500

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

$ =(\text{ Amount of loan })-(\text{ Sum of } 30 \text{ installments }) $

= 118000-73500= ₹ 44500

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

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

$\therefore$ Total distance covered by Ruchi for placing these flags

$ \begin{aligned} & =2 \times(4+8+12+\ldots+52) \\ & =2 \times \dfrac{13}{2}{2 \times 4+(13-1) \times(8-4)} \quad \quad \dfrac{\because \text{ Sum of } n \text{ terms of an AP }}{S_n=\dfrac{n}{2}[2 a+(n-1) d} \\ & .=2 \times \dfrac{13}{2}(8+12 \times 4) \quad \quad \quad \quad \because \text{ 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+\ldots+13 \text{ times }) \\ & =2 \times 13=26 m \end{aligned} $

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