If the sum of $n$ terms of the series $1 + \frac{4}{5} + \frac{7}{5^{2}} + \frac{10}{5^{3}} + \ldots$ is $l + \frac{15}{16}\left( 1 - \frac{1}{5^{n - 1}} \right) - \frac{(3n - 2)}{4(5^{n - 1})}$, then $l$ is

Question

Dr. Abubacker · Accepted Answer

(b)
Here, $a = 1,\ \ r = \frac{1}{5},\ \ d = 3$
$\because S_{n} = \frac{a}{1 - r} + \frac{dr(1 - r^{n - 1})}{(1 - r)^{2}} - \frac{\left\{ a + (n - 1)d ight\} r^{n}}{1 - r}$
$	herefore\ S_{n} = \frac{1}{1 - \frac{1}{5}} + \frac{3.\ \frac{1}{5}\left\lbrack 1 - \left( \frac{1}{5} ight)^{n - 1} ightbrack}{\left( 1 - \frac{1}{5} ight)^{2}} - \frac{\{ 1 + (n - 1)3\}}{1 - \frac{1}{5}}\ \left( \frac{1}{5} ight)^{n}$
$= \frac{5}{4} + \frac{15}{16}\left( 1 - \frac{1}{5^{n - 1}} ight) - \frac{1}{4}.\ \frac{(3n - 2)}{5^{n - 1}}$
But, $S_{n} = l + \frac{15}{16}\left( 1 - \frac{1}{5^{n - 1}} ight) - \frac{(3n - 2)}{{4.5}^{n - 1}}$
$	herefore\ l = \frac{5}{4}$

Q. If the sum of $n$ terms of the series $1 + \frac{4}{5} + \frac{7}{5^{2}} + \frac{10}{5^{3}} + \ldots$ is $l + \frac{15}{16}\left( 1 - \frac{1}{5^{n - 1}} \right) - \frac{(3n - 2)}{4(5^{n - 1})}$, then $l$ is

Explanation

Reviewed by Dr. Abubacker

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