Question

find the sum of infinity series 3+5.1/4+7.1/4^2....​

Answer 1

Answer:-

Given

AGP is 3 + 5 * 1/4 + 7 * 1/4²....

3 , 5 , 7 ... are in AP

a = 3

→ d = 5 - 3

→ d = 2

1 , 1/4 , 1/4² ... are in GP.

r = (1/4)/1

→ r = 1/4

We know that,

$\sf Sum\: of \:an\: infinite \:AGP\:– \:S_{\infty} = \dfrac{a}{1 - r} + \dfrac{dr}{(1 - r)^2 }$

Hence,

$\sf \implies \: S _{ \infty } = \dfrac{3}{1 - \dfrac{1}{4} } + \frac{ {2} \times \frac{1}{4} }{ \big(1 - \dfrac{1}{4} \big) ^{2} } \\ \\ \implies \sf \: S _{ \infty } = \dfrac{3}{ \frac{4 - 1}{4} } + \frac{ \frac{1}{2} }{ (\frac{4 - 1}{4}) ^{2} } \\ \\ \implies \sf \: S _{ \infty } = \frac{3}{3} \times 4 + \frac{1}{2} \times \frac{{4 }^{2} }{ {3}^{2} } \\ \\ \implies \sf \: S _{ \infty } = \: 4 + \frac{8}{9} \\ \\ \implies \sf \: S _{ \infty } = \: \frac{36 + 8}{9} \\ \\ \implies \sf \large S _{ \infty } = \frac{44}{9}$

Therefore, the sum of the given AGP is 44/9.

Answer 2

$\underline{\underline{\blue{Given:-}}}$

GP : - 3 + 5 × 1/4 + 7 ×1/4²....

• 3 , 5 , 7 ... are in AP

Here, a = 3

→ d = 5 - 3

→ d = 2

• 1 , 1/4 , 1/4² ... are in GP.

r = (1/4)/1

→ r = 1/4

We know :

$\sf{ \boxed{\pink{\underline{Sum\: of \:an\: infinite \:GP\:– \:S_{\infty} = a/1 - r+ dr/(1 - r)^2 }}}}$

Therefore,

$\begin{gathered}\sf ➝\: S _{ \infty } = \dfrac{3}{1 - \dfrac{1}{4} } + \frac{ {2} \times \frac{1}{4} }{ \big(1 - \dfrac{1}{4} \big) ^{2} } \\ \\ ➝\sf \: S _{ \infty } = \dfrac{3}{ \frac{4 - 1}{4} } + \frac{ \frac{1}{2} }{ (\frac{4 - 1}{4}) ^{2} } \\ \\ →\sf \: S _{ \infty } = \frac{3}{3} \times 4 + \frac{1}{2} \times \frac{{4 }^{2} }{ {3}^{2} } \\ \\ ➝\sf \: S _{ \infty } = \: 4 + \frac{8}{9} \\ \\ ➝\sf \: S _{ \infty } = \: \frac{36 + 8}{9} \\ \\ ➜ \sf \large S _{ \infty } = \frac{44}{9}\end{gathered}$

Hence, $\underline{\orange{The \: sum \: of given \: GP\: is \: 44/9.}}$

__________________________