Sequence and Series

Short Answer Type Questions

1. The first term of an AP is a and the sum of the first $p$ terms is zero, show that the sum of its next $q$ terms is $\frac{-a(p+q) q}{p-1}$.

Show Answer

Solution

Let the common difference of an AP is $d$. According to the question,

$S_{p}=0 \\ $

$\Rightarrow \frac{p}{2}[ 2 a+(p-1) d] =0 \quad [\because S_n = \frac{n}{2}{2a+(n-1)d}]$

$\Rightarrow 2 a+(p-1) d =0 \\ $

$\therefore d= \frac{-2 a}{p-1} \\ $

$\text { Now, sum of next } q \text { terms } =S_{p+q}-S_{p}=S_{p+q}-0 \\ $

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

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

$ = \frac{p+q}{2} \Big[2 a+(p-1) \cdot \frac{-2 a}{p-1}+\frac{q(-2 a)}{p-1}\Big] \\ $

$ =\frac{p+q}{2} \Big[2 a+(-2 a)-\frac{2 a q}{p-1}\Big] \\ $

$ = \frac{p+q}{2} \Big[\frac{-2 a q}{p-1}\Big] \\ $

$ = \frac{-a(p+q) q}{(p-1)} $

$\frac{p}{2}[2 a+(p-1) d]=0 \quad \because S_{n}=\frac{n}{2}{2 a+(n-1) d} $

2. A man saved ₹ 66000 in $20 yr$. In each succeeding year after the first year, he saved ₹ 200 more than what he saved in the previous year. How much did he save in the first year?

Show Answer

Solution

Let saved in first year ₹ a. Since, each succeeding year an increment ₹ 200 has made. So,it forms an AP whose

First term $=a$, common difference $(d)=200$ and $n=20 yr$

$ \begin{aligned} & \therefore \quad S_{20}=\frac{20}{2}[2 a+(20-1) d] \quad[\because S_{n}=\frac{n}{2}{2 a+(n-1) d}] \\ & \Rightarrow \quad 66000=10[2 a+19 d] \\ & \Rightarrow \quad 66000=20 a+190 d \\ & \Rightarrow \quad 66000=20 a+190 \times 200 \\ & \Rightarrow \quad 20 a=66000-38000 \\ & \Rightarrow \quad 20 a=28000 \\ & \therefore \quad a=\frac{28000}{20}=1400 \end{aligned} $

Hence, he saved ₹ 1400 in the first year.

3. A man accepts a position with an initial salary of ₹ 5200 per month. It is understood that he will receive an automatic increase of ₹ 320 in the very next month and each month thereafter.

(i) Find his salary for the tenth month.

(ii) What is his total earnings during the first year?

Show Answer

Solution

Since, the man get a fixed increment of ₹ 320 each month. Therefore, this forms an AP whose First term $=5200$ and Common difference $(d)=320$

(i) Salary for tenth month i.e., for $n=10$,

$ \begin{matrix} \Rightarrow & a_{10}=a+(n-1) d \\ \Rightarrow & a_{10}=5200+(10-1) \times 320 \\ \therefore & a_{10}=5200+9 \times 320 \\ \therefore & a_{10}=5200+2880 \\ a_{10}=8080 \end{matrix} $

(ii) Total earning during the first year.

In a year there are 12 month i.e., $n=12$,

$ \begin{aligned} S_{12} & =\frac{12}{2}[2 \times 5200+(12-1) 320] \\ & =6[10400+11 \times 320] \\ & =6[10400+3520]=6 \times 13920=83520 \end{aligned} $

4. If the $p$ th and $q$ th terms of a GP are $q$ and $p$ respectively, then show that its $(p+q)$ th term is $\frac{q^{p}}{p^{q}} \frac{1}{p-q}$.

Show Answer

Solution

Let the first term and common ratio of GP be a and $r$, respectively.

According to the question, $p$ th term $=q$

$\Rightarrow \quad a \cdot r^{p-1}=q \quad \quad …(i)$

and

$q$ th term $=p$

$\Rightarrow$

$ a r^{q-1}=p \quad \quad …(ii) $

On dividing Eq. (i) by Eq. (ii), we get

$ \begin{aligned} & \frac{a r^{p-1}}{a r^{q-1}}=\frac{q}{p} \\ & \Rightarrow \quad r^{p-1-q+1}=\frac{q}{p} \\ & \Rightarrow \quad r^{p-q}=\frac{q}{p} \Rightarrow r=\Big(\frac{q}p\Big)^{\frac{1}{p-q}} \end{aligned} $

On substituting the value of $r$ in Eq. (i), we get

$ \begin{aligned} & a \Big(\frac{q}p\Big)^{\frac{p-1}{p-q}}=q \Rightarrow a=\frac{q}{\Big(\frac{q}p\Big)^{\frac{p-1}{p-q}}}=q \cdot \Big(\frac{p}q\Big)^{\frac{p-1}{p-q}} \\ & \therefore \quad(p+q) \text { th term, } T_{p+q}=a \cdot r^{p+q-1}=q \cdot \frac{p}q^{\frac{p-1}{p-q}} \cdot(r)^{p+q-1} \\ &=q \cdot \Big(\frac{p}q\Big)^{\frac{p-1}{p-q}} \Big(\frac{q}p\Big)^{\frac{1}{p-q}}{ }^{p+q-1}=q \cdot \Big(\frac{p}q\Big)^{\frac{p-1}{p-q}} \frac{q}{\frac{p}{p}} \\ &=q \cdot \Big(\frac{p}q\Big)^{\frac{p-q-1}{p-q}} \frac{p}q^{\frac{-(p+q-1)}{p-q}}=q \cdot \Big(\frac{p}q\Big)^{\frac{p-1}{p-q}-\frac{(p+q-1)}{p-q}} \\ &=q \cdot \Big(\frac{p}q\Big)^{\frac{p-1-p-q+1}{p-q}}=q \cdot \Big(\frac{p}q\Big)^{\frac{-q}{p-q}} \\ & a=q \cdot \Big(\frac{p}q\Big)^{\frac{p-1}{p-q}} \end{aligned} $

Now, $(p+q)$ th term i.e., $a_{p+q}=a r^{p+q-1}$

$ \begin{aligned} & =q \cdot \Big(\frac{p}q\Big)^{\frac{p-1}{p-q}} \cdot \Big(\frac{q}p\Big)^{\frac{p+q-1}{p-q}} \\ & =q \cdot \frac{q \frac{p+q-1-p+1}{p-q}}{p \frac{p+q-1-p+1}{p-q}}=q \cdot \Big(\frac{q^{\frac{q}{p-q}}}{p^{\frac{q}{p-q}}}\Big) \end{aligned} $

5. A carpenter was hired to build 192 window frames. The first day he made five frames and each day, thereafter he made two more frames than he made the day before. How many days did it take him to finish the job?

Show Answer

Solution

Here, $a=5$ and $d=2$

Let he finished the job in $n$ days.

Then,

$ \begin{aligned} & S_{n}=192 \\ & S_{n}=\frac{n}{2}[2 a+(n-1) d] \end{aligned} $

$ \begin{matrix} \Rightarrow & 192=\frac{n}{2}[2 \times 5+(n-1) 2] \\ \Rightarrow & 192=\frac{n}{2}[10+2 n-2] \end{matrix} $

$\Rightarrow$ 192 $=\frac{n}{2}[8+2 n]$
$\Rightarrow$ 192 $=4 n+n^{2}$
$\Rightarrow$ $n^{2}+4 n-192$ $=0$
$\Rightarrow$ $(n-12)(n+16)$ $=0$
$\Rightarrow$ $n$ $n$ $=12,-16$
$n$ $=12$

6. The sum of interior angles of a triangle is $180^{\circ}$. Show that the sum of the interior angles of polygons with $3,4,5,6, \ldots$ sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon.

Show Answer

Solution

We know that, sum of interior angles of a polygon of side $n=(2 n-4) \times 90^{\circ}=(n-2) \times 180^{\circ}$ Sum of interior angles of a polygon with sides 3 is 180 .

Sum of interior angles of polygon with side $4=(4-2) \times 180^{\circ}=360^{\circ}$

Similarly, sum of interior angles of polygon with side $5,6,7 \ldots$ are $540^{\circ}, 720^{\circ}, 900^{\circ}, \ldots$

The series will be $180^{\circ}, 360^{\circ} 540^{\circ}, 720^{\circ}, 900^{\circ}, .$.

Here, $\quad a=180^{\circ}$

and $\quad d=360^{\circ}-180^{\circ}=180^{\circ}$

Since, common difference is same between two consecutive terms of the series.

So, it form an AP.

We have to find the sum of interior angles of a 21 sides polygon.

It means, we have to find the 19th term of the above series.

$ \begin{aligned} & \therefore \quad a_{19}=a+(19-1) d \\ & =180+18 \times 180=3420 \end{aligned} $

7. A side of an equilateral triangle is $20 cm$ long. A second equilateral triangle is inscribed in it by joining the mid-points of the sides of the first triangle. The process is continued as shown in the accompanying diagram. Find the perimeter of the sixth inscribed equilateral triangle.

Show Answer

Solution

Side of equilateral $\triangle A B C=20 cm$. By joining the mid-points of this triangle, we get another equilateral triangle of side equal to half of the length of side of $\triangle A B C$.

Continuing in this way, we get a set of equilateral triangles with side equal to half of the side of the previous triangle.

$\therefore \quad$ Perimeter of first triangle $=20 \times 3=60 cm$

Perimeter of second triangle $=10 \times 3=30 cm$

Perimeter of third triangle $=5 \times 3=15 cm$

Now, the series will be $60,30,15, \ldots$

Here,

$ a=60 $

$\therefore \quad r=\frac{30}{60}=\frac{1}{2} \quad \quad [\because \frac{\text { second term }}{\text { first term }}=r$]

We have, to find perimeter of sixth inscribed triangle. It is the sixth term of the series.

$ \begin{aligned} \therefore & a_6 & =a r^{6-1} \\ & =60 \times \frac{1}2^{5}=\frac{60}{32}=\frac{15}{8} cm & {[\because a_{n}=a r^{n-1}] } \end{aligned} $

8. In a potato race 20 potatoes are placed in a line at intervals of $4 m$ with the first potato $24 m$ from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?

Show Answer

Solution

According to the given information, we have following diagram.

Distance travelled to bring first potato $=24+24=2 \times 24=48 m$

Distance travelled to bring second potato $=2(24+4)=2 \times 28=56 m$

Distance travelled to bring third potato $=2(24+4+4)=2 \times 32=64 m$

Then, the series of distances are $48,56,64, \ldots$

Here,

$ \begin{aligned} & a=48 \\ & d=56-48=8 \end{aligned} $

$ \text { and } \quad n=20 $

To find the total distance that he run in bringing back all potatoes, we have to find the sum of 20 terms of the above series.

$ \begin{matrix} \therefore \quad S_{20} & =\frac{20}{2}[2 \times 48+19 \times 8] \quad \because S_{n}=\frac{n}{2}{2 a+(n-1) d} \\ & =10[96+152] & \\ & =10 \times 248=2480 m \end{matrix} $

9. In a cricket tournament 16 school teams participated. A sum of ₹ 8000 is to be awarded among themselves as prize money. If the last placed team is awarded ₹ 275 in prize money and the award increases by the same amount for successive finishing places, how much amount will the first place team receive?

Show Answer

Solution

Let the first place team got ₹ a.

Since, award money increases by the same amount for successive finishing places. Therefore series is an AP.

Let the constant amount be $d$.

Here, l = 275, n = 16 and $S_{16} = 8000$

$\therefore$ l = a+(n-)d

$\Rightarrow$ l = a+(16-1)(-d)

[we take common difference (−ve) because series is decreasing]

$\Rightarrow$ 275 = a-15d

and

$\Rightarrow \quad 8000=8[2 a+(16-1)(-d)]$

$\Rightarrow \quad 8000=8[2 a-15 d]$

$\Rightarrow \quad 1000=2 a-15 d$

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

$(2 a-15 d)-(a-15 d) =1000-275 \\ $

$\Rightarrow 2 a-15 d-a+15 d =725 \\ $

$\therefore a =725 $

Hence, first place team receive ₹ 725 .

10. If $a_1, a_2, a_3, \ldots, a_{n}$ are in AP, where $a_{i}>0$ for all $i$, show that

$ \frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}=\frac{n-1}{\sqrt{a_1}+\sqrt{a_{n}}} $

Show Answer

Solution

Since, $a_1, a_2, a_3, \ldots, a_{n}$ are in AP.

$ \begin{aligned} & \Rightarrow \quad a_2-a_1=a_3-a_2=\ldots=a_{n}-a_{n-1}=d \quad \text { [common difference] } \\ & \text { If } a_2-a_1=d \text {, then }(\sqrt{a_2})^{2}-(\sqrt{a_1})^{2}=d \\ & \Rightarrow \quad(\sqrt{a_2}-\sqrt{a_1})(\sqrt{a_2}+\sqrt{a_1})=d \\ & \Rightarrow \quad \frac{1}{\sqrt{a_1}+\sqrt{a_2}}=\frac{\sqrt{a_2}-\sqrt{a_1}}{d} \\ & \frac{1}{\sqrt{a_2}+\sqrt{a_3}}=\frac{\sqrt{a_3}-\sqrt{a_2}}{d} \\ & \frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}=\frac{\sqrt{a_{n}}-\sqrt{a_{n-1}}}{d} \end{aligned} $

On adding these terms, we get

$ \begin{aligned} & \frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}} \\ = & \frac{1}{d}[\sqrt{a_2}-\sqrt{a_1}+\sqrt{a_3}-\sqrt{a_2}+\ldots+\sqrt{a_{n}}-\sqrt{a_{n-1}}] \quad \text { [using above relations] } \\ = & \frac{1}{d}[\sqrt{a_{n}}-\sqrt{a_1}] \end{aligned} $

Again,

$ a_{n}=a_1+(n-1) d \quad[\because T_{n}=a+(n-1) d] $

$\Rightarrow$

$ a_{n}-a_1=(n-1) d $

$ \begin{matrix} \Rightarrow & (\sqrt{a_{n}})^{2}-(\sqrt{a_1})^{2}=(n-1) d \\ \Rightarrow & (\sqrt{a_{n}}-\sqrt{a_1})(\sqrt{a_{n}}+\sqrt{a_1})=(n-1) d \Rightarrow \sqrt{a_{n}}-\sqrt{a_1}=\frac{(n-1) d}{\sqrt{a_{n}}+\sqrt{a_1}} \end{matrix} $

On putting this value in Eq. (i), we get

$ \begin{gathered} \frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}} \\ =\frac{(n-1) d}{d(\sqrt{a_{n}}+\sqrt{a_1})}=\frac{n-1}{\sqrt{a_{n}}+\sqrt{a_1}} \end{gathered} $

11. Find the sum of the series

$ (3^{3}-2^{3})+(5^{3}-4^{3})+(7^{3}-6^{3})+\ldots \text { to (i) } n \text { terms. (ii) } 10 \text { terms. } $

Show Answer

Solution

Given series, $(3^{3}-2^{3})+(5^{3}-4^{3})+(7^{3}-6^{3})+$

$ =(3^{3}+5^{3}+7^{3}+\ldots)-(2^{3}+4^{3}+6^{3}+\ldots) $

Let $T_{n}$ be the $n$th term of the series (i),

then $T_{n}=(n.$th term of $.3^{3}, 5^{3}, 7^{3}, \ldots)-(n.$th term of $.2^{3}, 4^{3}, 6^{3}, \ldots)=(2 n+1)^{3}-(2 n)^{3}$

$ \begin{aligned} & =(2 n+1-2 n)[(2 n+1)^{2}+(2 n+1) 2 n+(2 n)^{2}] \quad[\because a^{3}-b^{3}=(a-b)(a^{2}+a b+b^{2})] \\ & =[4 n^{2}+1+4 n+4 n^{2}+2 n+4 n^{2}]=[12 n^{2}+6 n]+1 \end{aligned} $

(i) Let $S_{n}$ denote the sum of $n$ term of series (i). Then,

$ \begin{aligned} S_{n} & =\Sigma T_{n}=\Sigma(12 n^{2}+6 n) \\ & =12 \Sigma n^{2}+6 \Sigma n+\Sigma n \\ & =12 \cdot \frac{n(n+1)(2 n+1)}{6}+\frac{6 n(n+1)}{2}+n \\ & =2 n(n+1)(2 n+1)+3 n(n+1)+n \\ & =2 n(n+1)(2 n+1)+3 n(n+1)+n \\ & =(2 n^{2}+2 n)(2 n+1)+3 n^{2}+3 n+n \\ & =4 n^{3}+2 n^{2}+4 n^{2}+2 n+3 n^{2}+3 n+n \\ & =4 n^{3}+9 n^{2}+6 n \\ S_{10} & =4 \times(10)^{3}+9 \times(10)^{2}+6 \times 10 \\ & =4 \times 1000+9 \times 100+60 \\ & =4000+900+60=4960 \end{aligned} $

(ii) Sum of 10 terms,

12. Find the $r$ th term of an AP sum of whose first $n$ terms is $2 n+3 n^{2}$.

Show Answer

Solution

Given that, sum of $n$ terms of an AP,

$ \begin{aligned} S_{n} & =2 n+3 n^{2} \\ T_{n} & =S_{n}-S_{n-1} \\ & =(2 n+3 n^{2})-[2(n-1)+3(n-1)^{2}] \\ & =(2 n+3 n^{2})-[2 n-2+3(n^{2}+1-2 n)] \\ & =(2 n+3 n^{2})-(2 n-2+3 n^{2}+3-6 n) \\ & =2 n+3 n^{2}-2 n+2-3 n^{2}-3+6 n \\ \therefore \quad & =6 n-1 \\ \therefore \quad \text { rth term } T_{r} & =6 r-1 \end{aligned} $

Long Answer Type Questions

13. If $A$ is the arithmetic mean and $G_1, G_2$ be two geometric mean between any two numbers, then prove that $2 A=\frac{G_1^{2}}{G_2}+\frac{G_2^{2}}{G_1}$.

Show Answer

Solution

Let the numbers be $a$ and $b$.

Then,

$ A=\frac{a+b}{2} $

$ \Rightarrow \quad 2 A=a+b $

and $G_1, G_2$ be geometric mean between $a$ and $b$, then $a, G_1, G_2, b$ are in GP.

Let $r$ be the common ratio.

$ \begin{aligned} & \text { Then, } \\ & b=a r^{4-1} \\ & \Rightarrow \quad b=a r^{3} \Rightarrow \frac{b}{a}=r^{3} \\ & \therefore \quad r=\frac{b}a^{1 / 3} \end{aligned} $

$ \begin{aligned} & \text { Now, } \\ & G_1=a r=a \frac{b}a^{1 / 3} \\ & \text { and } \\ & G_2=a r^{2}=a \frac{b}a^{2 / 3} \\ & R H S=\frac{G_1^{2}}{G_2}+\frac{G_2^{2}}{G_1}=\frac{a \frac{b}a^{1 / 3^{2}}}{a \frac{b}{a}}+\frac{a \frac{b}a^{2 / 3}}{a \frac{b}{a}} \\ & =\frac{a^{2} \frac{b^{2 / 3}}{a}}{a \frac{b}{a}}+\frac{a^{2} \frac{b}{a}}{a \frac{b}{a}} \\ & =a+a \frac{b}{a}=a+b=2 A \\ & =LHS \end{aligned} $

14. If $\theta_1, \theta_2, \theta_3, \ldots, \theta_{n}$ are in AP whose common difference is $d$, show that $\sec \theta_1 \sec \theta_2+\sec \theta_2 \sec \theta_3+\ldots+\sec \theta_{n-1} \sec \theta_{n}=\frac{\tan \theta_{n}-\tan \theta_1}{\sin d}$.

Show Answer

Solution

Since, $\theta_1, \theta_2, \theta_3, \ldots, \theta_{n}$ are in AP.

$ \Rightarrow \quad \theta_2-\theta_1=\theta_3-\theta_2=\cdots=\theta_{n}-\theta_{n-1}=d $

$ \because r=\frac{b}a^{1 / 3} $

Now, we have to prove

$ \sec q_1 \sec q_2+\sec q_2 \sec q_3+\cdots+\sec q_{n-1} \sec \theta_{n}=\frac{\tan \theta_{n}-\tan \theta_1}{\sin d} $

or it can be written as

sind $[\sec \theta_1 \sec \theta_2+\sec \theta_2 \sec \theta_3+\cdots+\sec \theta_{n-1} \sec \theta_{n}]=\tan \theta_{n}-\tan \theta_1$

Now, taking only first term of LHS

$ \begin{aligned} \sin d \sec \theta_1 \sec \theta_2 & =\frac{\sin d}{\cos \theta_1 \cos \theta_2}=\frac{\sin (\theta_2-\theta_1)}{\cos \theta_1 \cos \theta_2} \\ & =\frac{\sin \theta_2 \cos \theta_1-\cos \theta_2 \sin \theta_1}{\cos \theta_1 \cos \theta_2} \\ & =\frac{\sin \theta_2 \cos \theta_1}{\cos \theta_1 \cos \theta_2}-\frac{\cos \theta_2 \sin \theta_1}{\cos \theta_1 \cos \theta_2}=\tan \theta_2-\tan \theta_1 \end{aligned} $

Similarly, we can solve other terms which will be $\tan \theta_3-\tan \theta_2, \tan \theta_4-\tan \theta_3, \cdots$

$ \begin{aligned} \therefore \quad LHS & =\tan \theta_2-\tan \theta_1+\tan \theta_3-\tan \theta_2+\cdots+\tan \theta_{n}-\tan \theta_{n-1} \\ & =-\tan \theta_1+\tan \theta_{n}=\tan \theta_{n}-\tan \theta_1 \\ & =\text { RHS } \quad \text { Hence proved. } \end{aligned} $

15. If the sum of $p$ terms of an AP is $q$ and the sum of $q$ terms is $p$, then show that the sum of $p+q$ terms is $-(p+q)$. Also, find the sum of first $p-q$ terms (where, $p>q$ ).

Show Answer

Solution

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

$ S_{p}=q $

$\Rightarrow$

$ \begin{aligned} \frac{p}{2}[2 a+(p-1) d] & =q \\ 2 a+(p-1) d & =\frac{2 q}{p} \end{aligned} $

and

$ S_{q}=p $

$ \begin{matrix} \Rightarrow & & \frac{q}{2}[2 a+(q-1) d] & =p \\ \Rightarrow & 2 a+(q-1) d & =\frac{2 p}{q} \end{matrix} $

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

$ 2 a+(p-1) d-2 a-(q-1) d =\frac{2 q}{p}-\frac{2 p}{q} \\ $

$\Rightarrow {[(p-1)-(q-1)] d} =\frac{2 q^{2}-2 p^{2}}{p q} \\ $

$\Rightarrow {[p-1-q+1] d} =\frac{2(q^{2}-p^{2})}{p q} \\ $

$\therefore d =\frac{-2(p+q)}{p q} $

On substituting the value of $d$ in Eq. (i), we get

$ \begin{aligned} & 2 a+(p-1) \quad \frac{-2(p+q)}{p q}=\frac{2 q}{p} \\ & \Rightarrow \quad 2 a=\frac{2 q}{p}+\frac{2(p+q)(p-1)}{p q} \\ & \Rightarrow \quad a=\frac{q}{p}+\frac{(p+q)(p-1)}{p q} \\ & \text { Now, } \quad S_{p+q}=\frac{p+q}{2}[2 a+(p+q-1) d] \\ &= \frac{p+q}{2} \frac{2 q}{p}+\frac{2(p+q)(p-1)}{p q}-\frac{(p+q-1) 2(p+q)}{p q} \\ &=(p+q) \frac{q}{p}+\frac{(p+q)(p-1)-(p+q-1)(p+q)}{p q} \\ &=(p+q) \frac{q}{p}+\frac{(p+q)(p-1-p-q+1)}{p q} \\ &=p+q \frac{q}{p}-\frac{p+q}{p}=(p+q) \frac{q-p-q}{p} \\ & S_{p+q}=-(p+q) \\ & S_{p-q}=\frac{p-q}{2}[2 a+(p-q-1) d] \end{aligned} $

$ \begin{aligned} & =\frac{p-q}{2} \frac{2 q}{p}+\frac{2(p+q)(p-1)}{p q}-\frac{(p-q-1) 2(p+q)}{p q} \\ & =(p-q) \frac{q}{p}+\frac{p+q(p-1-p+q+1)}{p q} \\ & =(p-q) \frac{q}{p}+\frac{(p+q) q}{p q} \\ & =(p-q) \frac{q}{p}+\frac{p+q}{p}=(p-q) \frac{(p+2 q)}{p} \end{aligned} $

16. If $p$ th, $q$ th and $r$ th terms of an AP and GP are both and $c$ respectively, then show that $a^{b-c} \cdot b^{c-a} \cdot c^{a-b}=1$.

Show Answer

Solution

Let $A, d$ are the first term and common difference of AP and $x, R$ are the first term and common ratio of GP, respectively.

According to the given condition,

and

$ \begin{aligned} A+(p-1) d & =a \\ A+(q-1) d & =b \\ A+(r-1) d & =c \\ a & =x R^{p-1} \\ b & =x R^{q-1} \\ c & =x R^{r-1} \end{aligned} $

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

$ d(p-1-q+1)=a-b $

$\Rightarrow \quad a-b=d(p-q)$

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

$ d(q-1-r+1)=b-c $

$\Rightarrow \quad b-c=d(q-r)$

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

$d(r-1-p+1) =c-a \\ $

$\Rightarrow c-a =d(r-p) $

Now, we have to prove $a^{b-c} b^{c-a} c^{a-b}=1$

$ \text { Taking LHS }=a^{b-c} b^{c-a} c^{a-b} $

Using Eqs. (iv), (v), (vi) and (vii), (viii), (ix),

$ \begin{aligned} LHS & =(x R^{p-1})^{d}(q-r)(x R^{q-1})^{d}(r-p)(x R^{r-1})^{d(p-q)} \\ & =x^{d(q-r)+d(r-p)+d(p-q)} R^{(p-1) d(q-r)+(q-1) d(r-p)+(r-1) d(p-q)} \\ & =x^{d(q-r+r-p+p-q)} \end{aligned} $

$ \begin{aligned} R^{d(p q-p r-q+r+q r-p q-r+p+r p-r q-p+q)} & =x^{0} R^{0}=1 \\ & =RHS \end{aligned} $

Objective Type Questions

17. If the sum of $n$ terms of an AP is given by $S_{n}=3 n+2 n^{2}$, then the common difference of the AP is

(a) 3

(b) 2

(c) 6

(d) 4

Show Answer

Solution

(d) Given, $S_{n}=3 n+2 n^{2}$

First term of the AP,

$ \begin{aligned} & \therefore \quad T_1=3 \times 1+2(1)^{2}=3+2=5 \\ & \text { and } \quad T_2=S_2-S_1 \\ & =[3 \times 2+2 \times(2)^{2}]-[3 \times 1+2 \times(1)^{2}] \\ & =14-5=9 \end{aligned} $

$\therefore$ Common difference $(d)=T_2-T_1=9-5=4$

18. If the third term of GP is 4 , then the product of its first 5 terms is

(a) $4^{3}$

(b) $4^{4}$

(c) $4^{5}$

(d) None of these

Show Answer

Solution

(c) It is given that, $T_3=4$

Let $a$ and $r$ the first term and common ratio, respectively.

Then,

$ a r^{2}=4 $

Product of first 5 terms $=a \cdot a r \cdot a r^{2} \cdot a r^{3} \cdot a r^{4}$

$ =a^{5} r^{10}=(a r^{2})^{5}=(4)^{5} $

19. If 9 times the 9 th term of an AP is equal to 13 times the 13 th term, then the 22 nd term of the AP is

(a) 0

(b) 22

(c) 198

(d) 220

Show Answer

Solution

(a) Let the first term be a and common difference be $d$.

According to the question, $\quad 9 \cdot T_9=13 \cdot T_{13}$

$ \begin{matrix} \Rightarrow & 9(a+8 d) & =13(a+12 d) \\ \Rightarrow & 9 a+72 d & =13 a+156 d \\ \Rightarrow & (9 a-13 a) & =156 d-72 d \\ \Rightarrow & -4 a & =84 d \\ \Rightarrow & a & =-21 d \\ \Rightarrow & a+21 d & =0 \\ \therefore & & \text { 22nd term i.e., } T_{22} & =[a+21 d] \\ & T_{22} & =0 \quad \text{[using Eq. (i)]} \end{matrix} $

20. If $x, 2 y$ and $3 z$ are in AP where the distinct numbers $x, y$ and $z$ are in GP, then the common ratio of the GP is

(a) 3

(b) $\frac{1}{3}$

(c) 2

(d) $\frac{1}{2}$

Show Answer

Solution

(b) Given, $x, 2 y$ and $3 z$ are in AP.

Then,

$ 2 y=\frac{x+3 z}{2} $

$\Rightarrow y =\frac{x+3 z}{4} \\ $

$\Rightarrow 4 y =x+3 z $

and $x, y, z$ are in GP

$ \begin{matrix} \text { Then, } & \frac{y}{x}=\frac{z}{y}=\lambda \\ \Rightarrow & y=x \lambda \text { and } z=\lambda y=\lambda^{2} x \end{matrix} $

On substituting these values in Eq. (i), we get

$ 4(x \lambda) =x+3(\lambda^{2} x) \\ $

$\Rightarrow 4 \lambda x =x+3 \lambda^{2} x \\ $

$\Rightarrow 4 \lambda =1+3 \lambda^{2} \\ $

$\Rightarrow 3 \lambda^{2}-4 \lambda+1 =0 \\ $

$\Rightarrow (3 \lambda-1)(\lambda-1) =0 \\ $

$\therefore \lambda =\frac{1}{3}, \lambda=1 $

21. If in an AP, $S_{n}=q n^{2}$ and $S_{m}=q m^{2}$, where $S_{r}$ denotes the sum of $r$ terms of the AP, then $S_{q}$ equals to

(a) $\frac{q^{3}}{2}$

(b) $m n q$

(c) $q^{3}$

(d) $(m+n) q^{2}$

Show Answer

Solution

(c) Given, $S_{n}=q n^{2}$ and $S_{m}=q m^{2}$

$ \begin{aligned} & \therefore \quad S_1=q, S_2=4 q, S_3=9 q \text { and } S_4=16 q \\ & \text { Now, } \quad T_1=q \\ & \therefore \quad T_2=S_2-S_1=4 q-q=3 q \\ & T_3=S_3-S_2=9 q-4 q=5 q \\ & T_4=S_4-S_3=16 q-9 q=7 q \end{aligned} $

So, the series is $q, 3 q, 5 q, 7 q, \ldots$

$ \begin{aligned} & \text { Here, } \quad a=q \text { and } d=3 q-q=2 q \\ & \therefore \quad S_{q}=\frac{q}{2}[2 \times q+(q-1) 2 q] \\ & =\frac{q}{2} \times[2 q+2 q^{2}-2 q]=\frac{q}{2} \times 2 q^{2}=q^{3} \end{aligned} $

22. Let $S_{n}$ denote the sum of the first $n$ terms of an AP, if $S_{2 n}=3 S_{n}$, then $S_{3 n}: S_{n}$ is equal to

(a) 4

(b) 6

(c) 8

(d) 10

Show Answer

Solution

(b) Let first term be a and common difference be $d$.

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

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

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

$S_{3 n}=\frac{3 n}{2}[2 a+(3 n-1) d]$

According to the question, $S_{2 n}=3 S_{n}$

$ \begin{aligned} \Rightarrow & & n[2 a+(2 n-1) d] & =3 \frac{n}{2}[2 a+(n-1) d] \\ \Rightarrow & & 4 a+(4 n-2) d & =6 a+(3 n-3) d \\ \Rightarrow & & -2 a+(4 n-2-3 n+3) d & =0 \\ \Rightarrow & & -2 a+(n+1) d & =0 \\ \Rightarrow & & d & =\frac{2 a}{n+1} \end{aligned} $

Now,

$ \begin{aligned} \frac{S_{3 n}}{S_{n}} & =\frac{\frac{3 n}{2}[2 a+(3 n-1) d]}{\frac{n}{2}[2 a+(n-1) d]}=\frac{6 a+(9 n-3) \frac{2 a}{n+1}}{2 a+(n-1) \frac{2 a}{n+1}} \\ & =\frac{6 a n+6 a+18 a n-6 a}{2 a n+2 a+2 a n-2 a} \\ & =\frac{24 a n}{4 a n}=\frac{S_{3 n}}{S_{n}}=6 \end{aligned} $

23. The minimum value of $4^{x}+4^{1-x}, x \in R$ is

(a) 2

(b) 4

(c) 1

(d) 0

Show Answer

Solution

(b) We know that,

$ AM \geq GM $

$ \begin{matrix} \Rightarrow & \frac{4^{x}+4^{1-x}}{2} \geq \sqrt{4^{x} \cdot 4^{1-x}} \\ \Rightarrow & 4^{x}+4^{1-x} \geq 2 \sqrt{4} \\ \Rightarrow & 4^{x}+4^{1-x} \geq 2 \cdot 2 \\ \Rightarrow & 4^{x}+4^{1-x} \geq 4 \end{matrix} $

24. Let $S_{n}$ denote the sum of the cubes of the first $n$ natural numbers and $s_{n}$ denote the sum of the first $n$ natural numbers, then $\sum_{r=1}^{n} \frac{S_{r}}{S_4}$ equals to (a) $\frac{n(n+1)(n+2)}{6}$ (b) $\frac{n(n+1)}{2}$ (c) $\frac{n^{2}+3 n+2}{2}$ (d) None of these

Show Answer

Solution

(a) $\quad \sum_{r=1}^{n} \frac{S_{r}}{S_{r}}=\frac{S_1}{S_1}+\frac{S_2}{S_2}+\frac{S_3}{S_3}+\ldots+\frac{S_{n}}{S_{n}}$

Let $T_{n}$ be the $n$th term of the above series.

$ \begin{aligned} \therefore \quad T_{n} & =\frac{S_{n}}{S_{n}}=\frac{\frac{n(n+1)^{2}}{2}}{\frac{n(n+1)}{2}} \\ & =\frac{n(n+1)}{2}=\frac{1}{2}[n^{2}+n] \end{aligned} $

$\therefore$ Sum of the above series $=\Sigma T_{n}=\frac{1}{2}[\Sigma n^{2}+\Sigma n]$

$ \begin{aligned} & =\frac{1}{2} \frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}=\frac{1}{2} \cdot \frac{n(n+1)}{2} \frac{(2 n+1)}{3}+1 \\ & =\frac{1}{4} n(n+1) \frac{2 n+1+3}{3}=\frac{1}{4 \times 3} n(n+1)(2 n+4) \\ & =\frac{1}{12} n(n+1)(2 n+4)=\frac{1}{6} n(n+1)(n+2) \end{aligned} $

25. If $t_{n}$ denotes the $n$th term of the series $2+3+6+11+18+\ldots$, then $t_{50}$ is

(a) $49^{2}-1$

(b) $49^{2}$

(c) $50^{2}+1$

(d) $49^{2}+2$

Show Answer

Solution

(d) Let $S_{n}$ be sum of the series $2+3+6+11+18+\ldots+t_{50}$.

$ \begin{matrix} \therefore & S_{n}=2+3+6+11+18+\ldots+t_{50} \\ \text { and } & S_{n}=0+2+3+6+11+18+\ldots+t_{49}+t_{50} \end{matrix} $

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

$ \begin{aligned} & 0 & =2+1+3+5+7+\cdots+t_{50} \\ \Rightarrow & t_{50} & =2+1+3+5+7+\cdots \text { upto } 49 \text { terms } \\ \therefore & t_{50} & =2+[1+3+5+7+\cdots \text { upto } 49 \text { terms }] \\ & & =2+\frac{49}{2}[2 \times 1+48 \times 2] \\ & & =2+\frac{49}{2} \times[2+96] \\ & & =2+[49+49 \times 48] \\ & & =2+49 \times 49=2+(49)^{2} \end{aligned} $

26. The lengths of three unequal edges of a rectangular solid block are in GP. If the volume of the block is $216 cm^{3}$ and the total surface area is $252 cm^{2}$, then the length of the longest edge is

(a) $12 cm$

(b) $6 cm$

(c) $18 cm$

(d) $3 cm$

Show Answer

Solution

(a) Let the length, breadth and height of rectangular solid block is $\frac{a}{r}$, a and ar, respectively.

$ \begin{aligned} & \therefore \quad \text { Volume }=\frac{a}{r} \times a \times a r=216 cm^{3} \\ & \Rightarrow \quad a^{3}=216 \Rightarrow a^{3}=6^{3} \\ & \therefore \quad a=6 \\ & \text { Surface area }=2 \frac{a^{2}}{r}+a^{2} r+a^{2}=252 \\ & \Rightarrow \quad 2 a^{2} \frac{1}{r}+r+1=252 \\ & \Rightarrow \quad 2 \times 36 \frac{1+r^{2}+r}{r}=252 \\ & \Rightarrow \quad \frac{1+r^{2}+r}{r}=\frac{252}{2 \times 36} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 1+r^{2}+r=\frac{126}{36} r \Rightarrow 1+r^{2}+r=\frac{21}{6} r \\ & \Rightarrow \quad 6+6 r^{2}+6 r=21 r \Rightarrow 6 r^{2}-15 r+6=0 \\ & \Rightarrow \quad 2 r^{2}-5 r+2=0 \Rightarrow(2 r-1)(r-2)=0 \\ & \therefore \quad r=\frac{1}{2}, 2 \\ & \text { For } r=\frac{1}{2}: \quad \text { Length }=\frac{a}{r}=\frac{6 \times 2}{1}=12 \\ & \text { Breadth }=a=6 \\ & \text { Height }=a r=6 \times \frac{1}{2}=3 \\ & \text { For } r=2: \quad \text { Length }=\frac{a}{r}=\frac{6}{2}=3 \\ & \text { Breadth }=a=6 \\ & \text { Height }=a r=6 \times 2=12 \end{aligned} $

Fillers

27. If $a, b$ and $c$ are in GP, then the value of $\frac{a-b}{b-c}$ is equal to ……

Show Answer

Solution

Given that, $a, b$ and $c$ are in GP.

$ \begin{aligned} & \text { Then, } \\ & \frac{b}{a}=\frac{c}{b}=r \\ & \Rightarrow \quad b=a r \Rightarrow c=b r \\ & \begin{matrix} \Rightarrow & \frac{a-b}{b-c}=\frac{a-a r}{a r-b r}=\frac{a(1-r)}{r(a-b)}=\frac{a(1-r)}{r(a-a r)} \end{matrix} \\ & =\frac{a(1-r)}{a r(1-r)}=\frac{1}{r} \\ & \therefore \quad \frac{a-b}{b-c}=\frac{1}{r}=\frac{a}{b} \text { or } \frac{b}{c} \end{aligned} $

28. The sum of terms equidistant from the beginning and end in an AP is equal to ……

Show Answer

Solution

Let AP be $a, a+d, a+2 d \cdots a+(n-1) d$

$ \begin{aligned} & \therefore \quad a_1+a_{n}=a+a+(n-1) d \\ & =2 a+(n-1) d \\ & a_2+a_{n-1}=(a+d)+[a+(n-2) d] \\ & =2 a+(n-1) d \\ & a_2+a_{n-1}=a_1+a_{n} \\ & a_3+a_{n-2}=(a+2 d)+[a+(n-3) d] \\ & =2 a+(n-1) d \\ & =a_1+a_{n} \end{aligned} $

an $AP$ is equal to [first term +last term].

29. The third term of a GP is 4 , the product of the first five terms is …..

Show Answer

Solution

It is given that, $T_3=4$

Let $a$ and $r$ the first term and common ration, respectively.

Then,

$ a r^{2}=4 $

Product of first 5 terms $=a r \cdot a r \cdot a r^{2} \cdot a r^{3} \cdot a r^{4}$

$ =a^{5} r^{10}=(a r^{2})^{5}=(4)^{5} $

[using Eq. (i)]

True/False

30. Two sequences cannot be in both AP and GP together.

Show Answer

Solution

False

Consider an AP $a, a+d, a+2 d, \ldots$

Now,

$ \frac{a_2}{a_1}=\frac{a+d}{a} \neq \frac{a+2 d}{a+d} $

Thus, AP is not a GP.

31. Every progression is a sequence but the converse, i.e., every sequence is also a progression need not necessarily be true.

Show Answer

Solution

True

Consider the progression $a, a+d, a+2 d, \ldots$

and sequence of prime number $2,3,5,7,11, \ldots$

Clearly, progression is a sequence but sequence is not progression because it does not follow a specific pattern.

32. Any term of an AP (except first) is equal to half the sum of terms which are equidistant from it.

Show Answer

Solution

True

Consider an AP $a, a+d, a+2 d, \ldots$

Now,

$ \begin{aligned} a_2+a_4 & =a+d+a+3 d \\ & =2 a+4 d=2 a_3 \\ a_3 & =\frac{a_2+a_4}{2} \\ \frac{a_3+a_5}{2} & =\frac{a+2 d+a+4 d}{2}=\frac{2 a+6 d}{2} \\ & =a+3 d=a_4 \end{aligned} $

$ \Rightarrow \quad a_3=\frac{a_2+a_4}{2} $

Again

Hence, the statement is true.

33. The sum or difference of two GP, is again a GP.

Show Answer

Solution

False

Let two GP are $a, a r_1, a r_1^{2}, a r_2^{3}, \ldots$ and $b, b r_2, b r_2^{2}, b r_2^{3}, \ldots$

Now, sum of two GP $a+b,(a r_1+b r_2),(a r_1^{2}+b r_2^{2}), \ldots$

Now, $\quad \frac{T_2}{T_1}=\frac{a r_1+b r_2}{a+b}$ and $\frac{T_3}{T_2}=\frac{a r_1^{2}+b r_2^{2}}{a r_1+b r_2}$

$\therefore \quad \frac{T_2}{T_1} \neq \frac{T_3}{T_2}$

Again, difference of two GP is $a-b, a r_1-b r_2, a r_1^{2}-b r_2^{2}, \ldots$

Now,

$ \frac{T_2}{T_1}=\frac{a r_1-b r_2}{a-b} \text { and } \frac{T_3}{T_2}=\frac{a r_1^{2}-b r_2^{2}}{a r_1-b r_2} $

$ \therefore \quad \frac{T_2}{T_1} \neq \frac{T_3}{T_2} $

So, the sum or difference of two GP is not a GP. Hence, the statement is false.

34. If the sum of $n$ terms of a sequence is quadratic expression, then it always represents an AP.

Show Answer

Solution

False

Let

$\begin{aligned} S_n & =a n^2+b n+c \\ S_1 & =a+b+c \\ a_1 & =a+b+c \\ S_2 & =4 a+2 b+c \\ a_2 & =S_2-S_1 \\ & =4 a+4 b+c-(a+b+c)=3 a+b \\ S_3 & =9 a+3 b+c \\ a_3 & =S_3-S_2=5 a+b \end{aligned}$

Now,$\quad a_2-a_1=(3 a+b)-(a+b+c)=2 a-c$

$\quad a_3-a_2=(5 a+b)-(3 a+b)=2 a$

Now,$\quad a_2-a_1 \neq a_3-a_2$

Hence, the statement is false.

Matching The Columns

35. Match the following.

Column I Column II
(i) $4,1, \frac{1}{4}, \frac{1}{16}$ (a) AP
(ii) $2,3,5,7$ (b) Sequence
(iii) $13,8,3,-2,-7$ (c) GP
Show Answer

Solution

(i) $4,1, \frac{1}{4}, \frac{1}{16}$

$\Rightarrow$

$ \frac{T_2}{T_1}=\frac{1}{4} \Rightarrow \frac{T_3}{T_2}=\frac{1}{4} \Rightarrow \frac{T_4}{T_3}=\frac{1 / 16}{1 / 4}=\frac{1}{4} $

Hence, it is a GP.

(ii) $2,3,5,7$

$ \begin{matrix} \because & T_2-T_1=3-2=1 \\ \because & T_3-T_2=5-3=2 \\ & T_2-T_1 \neq T_3-T_2 \end{matrix} $

Hence, it is not an AP.

Again,

$ \frac{T_2}{T_1}=3 / 2 \Rightarrow \frac{T_3}{T_2}=5 / 3 $

$ \because \quad \frac{T_2}{T_1} \neq \frac{T_3}{T_2} $

It is not a GP.

Hence, it is a sequence.

(iii) $13,8,3,-2,-7$

$\because \quad T_2-T_1=T_3-T_2$

Hence, it is an AP.

$ \begin{aligned} & T_2-T_1=8-13=-5 \\ & T_3-T_2=3-8=-5 \\ & T_2-T_1=T_3-T_2 \end{aligned} $

36. Match the following.

Column I Column II
(i) $1^{2}+2^{2}+3^{2}+\cdots+n^{2}$ (a) $\frac{n(n+1)^{2}}{2}$
(ii) $1^{3}+2^{3}+3^{3}+\cdots+n^{3}$ (b) $n(n+1)$
(iii) $2+4+6+\cdots+2 n$ (c) $\frac{n(n+1)(2 n+1)}{6}$
(iv) $1+2+3+\cdots+n$ (d) $\frac{n(n+1)}{2}$
Show Answer

Solution

(i) $1^{2}+2^{2}+3^{2}+\cdots+n^{2}$

Consider the identity, $(k+1)^{3}-k^{3}=3 k^{2}+3 k+1$

On putting $k=1,2,3, \ldots,(n-1), n$ successively, we get

$ \begin{aligned} & 2^{3}-1^{3}=3 \cdot 1^{2}+3 \cdot 1+1 \\ & 3^{3}-2^{3}=3 \cdot 2^{2}+3 \cdot 2+1 \\ & 4^{3}-3^{3}=3 \cdot 3^{2}+3 \cdot 3+1 \end{aligned} $

$ \begin{aligned} & n^{3}-(n-1)^{3}=3 \cdot(n-1)^{2}+3 \cdot(n-1)+1 \\ & (n+1)^{3}-n^{3}=3 \cdot n^{2}+3 \cdot n+1 \end{aligned} $

Adding columnwise, we get

$n^{3}+3 n^{2}+3 n=3 \quad \sum_{r=1}^{n} r^{2}+3 \frac{n(n+1)}{2}+n \quad \because \sum_{r=1}^{n} r^{2}=\frac{n(n+1)}{2} \\ $

$\Rightarrow \quad 3 \sum_{r=1}^{n} r^{2}=n^{3}+3 n^{2}+3 n-\frac{3 n(n+1)}{2}+n \\ $

$\Rightarrow \quad \sum_{r=1}^{n} r^{2}=\frac{2 n^{3}+3 n^{2}+n}{2}=\frac{n(n+1)(2 n+1)}{2} \\ $

$ \Rightarrow \quad \sum_{r=1}^{n} r^{2}=\frac{n(n+1)(2 n+1)}{6} $

Hence, $\sum_{r=1}^{n} r^{2}=1^{2}+2^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}$ (ii) $1^{3}+2^{3}+3^{3}+\cdots+n^{3}$

Consider the identity $(k+1)^{4}-k^{4}=4 k^{3}+6 k^{2}+4 k+1$

On putting $k=1,2,3, \cdots(n-1), n$ successively, we get

$ \begin{aligned} & 2^{4}-1^{4}=4 \cdot 1^{3}+6 \cdot 1^{2}+4 \cdot 1+1 \\ & 3^{4}-2^{4}=4 \cdot 2^{3}+6 \cdot 2^{2}+4 \cdot 2+1 \\ & 4^{4}-3^{4}=4 \cdot 3^{3}+6 \cdot 3^{2}+4 \cdot 3+1 \end{aligned} $

$ \begin{aligned} & n^{4}-(n-1)^{4}=4(n-1)^{3}+6(n-1)^{2}+4(n-1)+1 \\ & (n+1)^{4}-n^{4}=4 \cdot n^{3}+6 \cdot n^{2}+4 \cdot n+1 \end{aligned} $

Adding columnwise, we get

$(n+1)^{4}-1^{4}=4 \cdot(1^{3}+2^{3}+\cdots+n^{3})+6(1^{2}+2^{2}+3^{3}+\cdots+n^{2}) \\ $

$+4(1+2+3+\cdots+n)+(1+1+\cdots+1) n \text { terms } \\ $

$\Rightarrow n^{4}+4 n^{3}+6 n^{2}+4 n=4 \sum_{r=1}^{n} r^{3}+6 \sum_{r=1}^{n} r^{2}+4 \sum_{r=1}^{n} r+n \\ $

$\Rightarrow n^{4}+4 n^{3}+6 n^{2}+4 n=4 \sum_{r=1}^{n} r^{3}+6 \frac{n(n+1)(2 n+1)}{6}+4 \frac{n(n+1)}{2}+n \\ $

$\Rightarrow \sum_{r=1}^{n} r^{3}=\frac{n^{2}(n+1)^{2}}{4} \\ $

$\Rightarrow \sum_{r=1}^{n} r^{3}=\frac{n(n+1)}{2}=\sum_{r=1}^{n} r^{2} \\ $

$\text { Hence, } \sum_{r=1}^{n} r^{3}=1^{3}+2^{3}+\cdots+n^{3}=\frac{n(n+1)^{2}}{2}=\sum_{r=1}^{n} r^{2} $

(iii)

$ \begin{aligned} 2+4+6+\cdots+2 n & =2[1+2+3+\cdots+n] \\ & =2 \times \frac{n(n+1)}{2}=n(n+1) \end{aligned} $

(iv) Let

$ S_{n}=1+2+3+\cdots+n $

Clearly, it is an arithmetic series with first term, $a=1$,

common difference,

$d=1$

and

last term $=n$

$ S_{n}=\frac{n}{2}(1+n)=\frac{n(n+1)}{2} $

Hence, $\quad 1+2+3+\cdots+n=\frac{n(n+1)}{2}$.



Mock Test for JEE

NCERT Chapter Video Solution

Dual Pane