Question

Find the common ratio of an infinite G.P. whose each term is five times the sum of its succeeding terms,explain please

Answer 1

The given infinite Geometric progression has each of its terms equal to five times sum of its succeeding terms.

The sum of infinite series of a GP is

$S = \frac{a}{1-r}$   , where |r| < 1

$S_n = \frac{a(1-r^n)}{1-r}$

$ar^{n-1} = 5(S-S_n).....given$

$ar^{n-1} = 5[\frac{a}{1-r}-\frac{a(1-r^n)}{1-r}]$

$ar^{n-1} = 5[\frac{a-a+ar^n}{1-r}]$

$ar^{n-1} = 5[\frac{ar^n}{1-r}]$

$\frac{ar^{n-1}}{ar^n}= \frac{5}{1-r}$

$r^{-1} = \frac{5}{1-r}$

$\frac{1}{r} = \frac{5}{1-r}$

$5r = 1-r$

$6r = 1$

$r = \frac{1}{6}$

Hence the common ratio is 1/6

HOPE THIS HELPS!!