Question

For a G.P.
i) if a = 2, r = 3, $\rm S_{n}$ = 242, find n.
ii) if $\rm S_{3}$ = 125, $\rm S_{6}$ = 152, find r.

Answer 1

Solution:

Sum of n terms of a G.P. is given by

$S_{n} = \frac{a( {r}^{n} - 1)}{r - 1} \: \: \: r > 1 \\ \\ S_{n} = \frac{a( 1 - {r}^{n} )}{1 - r } \: \: \: r < 1$
i) if a = 2, r = 3, $\rm S_{n}$ = 242, find n.

Sum of n terms is given by first formula

$S_{n} = \frac{a( {r}^{n} - 1)}{r - 1} \: \: \: r > 1$
$242 = \frac{2({3}^{n} - 1) }{3 - 1} \\ \\ 242 = \frac{2( {3}^{n} - 1) }{2} \\ \\242 = ({3)}^{n} - 1\\ \\ 243 = ( {3)}^{n} \\ \\ {3}^{5} = {3}^{n} \\ \\ n = 5$

2) if $\rm S_{3}$ = 125, $\rm S_{6}$ = 152, find r.
$S_{3}= 125 \\ \\ 125 = \frac{a({r}^{3} - 1)}{r - 1} ...eq1$
$S_{6}= 152 \\ \\ 152 = \frac{a({r}^{6} - 1)}{r - 1} \\ \\ 152= \frac{a( {r}^{3} - 1)( {r}^{3} +1 )}{r - 1} \\ \\ place \: value \: from \: eq1 \\ \\ 152= 125( {r}^{3} + 1) \\ \\ \frac{152}{125} - 1 = {r}^{3} \\ \\ {r}^{3} = \frac{27}{125} \\ \\ {r}^{3} = \bigg( { \frac{3}{5} }\bigg)^{3} \\ \\ so \\ \\ r = \frac{3}{5}$
Hope it helps you.