Question

CORRECT ANSWER WILL GO FOR BRAINLIEST

Solve question number 36​

Answer 1

$\textsf{\large{\underline{Solution 1}:}}$

For the given AP, let us assume that:

$\rm:\longmapsto a=First\ term.$

$\rm:\longmapsto d=Common\ difference.$

Therefore:

$\rm:\longmapsto a_{n}=a+(n-1)d$

$\rm:\longmapsto S_{n}=\dfrac{n}{2}[2a+(n-1)d]$

Now, according to the question:

$\rm:\longmapsto S_{2n}=3\: S_{n}$

$\rm:\longmapsto \dfrac{2n}{2}[2a+(2n-1)d]=3\cdot \dfrac{n}{2}[2a+(n-1)d]$

$\rm:\longmapsto n[2a+(2n-1)d]=3\cdot \dfrac{n}{2}[2a+(n-1)d]$

$\rm:\longmapsto [2a+(2n-1)d]=\dfrac{3}{2}[2a+(n-1)d]$

$\rm:\longmapsto 2[2a+(2n-1)d]=3[2a+(n-1)d]$

$\rm:\longmapsto 4a+4dn-2d=6a+3nd-3d$

$\rm:\longmapsto [6a-4a+3nd-4nd-3d+2d=0$

$\rm:\longmapsto 2a-(n+1)d=0$

$\rm:\longmapsto 2a=(n+1)d$

Now, taking LHS, we get:

$\rm=\dfrac{S_{3n}}{S_{n}}$

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

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

$\rm=\dfrac{3[(n+1)d+(3n-1)d]}{[(n+1)d+(n-1)d]}$

$\rm=\dfrac{3[nd+d+3nd-d]}{[nd+d+nd-d]}$

$\rm=\dfrac{3\cdot 4nd}{2nd}$

$\rm=3\times 2$

$\rm=6$

Therefore:

$\rm:\longmapsto\dfrac{S_{3n}}{S_{n}}=6$

Hence Proved!

$\textsf{\large{\underline{Solution 2}:}}$

Given AP:  -6, -11/2, -5, . . .

Here:

$\rm:\longmapsto a=-6$

$\rm:\longmapsto d=\dfrac{-11}{2}-(-6)=\dfrac{-11}{2}+6=\dfrac{1}{2}$

Let number of terms required be n.

Therefore:

$\rm:\longmapsto S_{n}=-25$

$\rm:\longmapsto \dfrac{n}{2}[2a+(n-1)d]=-25$

$\rm:\longmapsto \dfrac{n}{2}[2\times(-6)+(n-1)\times0.5]=-25$

$\rm:\longmapsto \dfrac{n}{2}[-12+0.5n-0.5]=-25$

$\rm:\longmapsto \dfrac{n}{2}[-12.5+0.5n]=-25$

$\rm:\longmapsto \dfrac{n}{2}\bigg(\dfrac{-25}{2}+\dfrac{n}{2}\bigg)=-25$

$\rm:\longmapsto \dfrac{n(n-25)}{4}=-25$

$\rm:\longmapsto n(n-25)=-100$

$\rm:\longmapsto n^{2}-25n+100=0$

$\rm:\longmapsto n^{2}-20n-5n+100=0$

$\rm:\longmapsto n(n-20)-5(n-20)=0$

$\rm:\longmapsto (n-5)(n-20)=0$

Therefore:

$\rm:\longmapsto n=5, 20$

Which is our required answer.

Answer 2

Answer:

$[tex]\textsf{\large{\underline{Solution 1}:}}$

For the given AP, let us assume that:

$\rm:\longmapsto a=First\ term.$

$\rm:\longmapsto d=Common\ difference.$

Therefore:

$\rm:\longmapsto a_{n}=a+(n-1)d$

$\rm:\longmapsto S_{n}=\dfrac{n}{2}[2a+(n-1)d]$

Now, according to the question:

$\rm:\longmapsto S_{2n}=3\: S_{n}$

$\rm:\longmapsto \dfrac{2n}{2}[2a+(2n-1)d]=3\cdot \dfrac{n}{2}[2a+(n-1)d]$

$\rm:\longmapsto n[2a+(2n-1)d]=3\cdot \dfrac{n}{2}[2a+(n-1)d]$

$\rm:\longmapsto [2a+(2n-1)d]=\dfrac{3}{2}[2a+(n-1)d]$

$\rm:\longmapsto 2[2a+(2n-1)d]=3[2a+(n-1)d]$

$\rm:\longmapsto 4a+4dn-2d=6a+3nd-3d$

$\rm:\longmapsto [6a-4a+3nd-4nd-3d+2d=0$

$\rm:\longmapsto 2a-(n+1)d=0$

$\rm:\longmapsto 2a=(n+1)d$

Now, taking LHS, we get:

$\rm=\dfrac{S_{3n}}{S_{n}}$

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

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

$\rm=\dfrac{3[(n+1)d+(3n-1)d]}{[(n+1)d+(n-1)d]}$

$\rm=\dfrac{3[nd+d+3nd-d]}{[nd+d+nd-d]}$

$\rm=\dfrac{3\cdot 4nd}{2nd}$

$\rm=3\times 2$

$\rm=6$

Therefore:

$\rm:\longmapsto\dfrac{S_{3n}}{S_{n}}=6$

Hence Proved!

$\textsf{\large{\underline{Solution 2}:}}$

Given AP:  -6, -11/2, -5, . . .

Here:

$\rm:\longmapsto a=-6$

$\rm:\longmapsto d=\dfrac{-11}{2}-(-6)=\dfrac{-11}{2}+6=\dfrac{1}{2}$

Let number of terms required be n.

Therefore:

$\rm:\longmapsto S_{n}=-25$

$\rm:\longmapsto \dfrac{n}{2}[2a+(n-1)d]=-25$

$\rm:\longmapsto \dfrac{n}{2}[2\times(-6)+(n-1)\times0.5]=-25$

$\rm:\longmapsto \dfrac{n}{2}[-12+0.5n-0.5]=-25$

$\rm:\longmapsto \dfrac{n}{2}[-12.5+0.5n]=-25$

$\rm:\longmapsto \dfrac{n}{2}\bigg(\dfrac{-25}{2}+\dfrac{n}{2}\bigg)=-25$

$\rm:\longmapsto \dfrac{n(n-25)}{4}=-25$

$\rm:\longmapsto n(n-25)=-100$

$\rm:\longmapsto n^{2}-25n+100=0$

$\rm:\longmapsto n^{2}-20n-5n+100=0$

$\rm:\longmapsto n(n-20)-5(n-20)=0$

$\rm:\longmapsto (n-5)(n-20)=0$

Therefore:

$\rm:\longmapsto n=5, 20$

Which is our required answer.[/tex]