Question

For a G.P.
i) If a = 1, r = -3/2 find $\rm S_{5}$
ii) If $\rm S_{5}$ =1023, r = 4, find a.

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 = 1, r = -3/2 find $\rm S_{5}$

here a= 1

$r= \frac{ - 3}{2}$

Sum of n terms is given by second formula

$S_{n} = \frac{a( 1 - {r}^{n} )}{1 - r } \: \: \: r < 1 \\ \\ S_{5} = \frac{1( 1 - {( \frac{-3}{2})}^{5} )}{1 - \frac{-3}{2} } \: \: \: r < 1 \\ \\ S_{5}= \frac{1(1 - (\frac{-243}{32}) }{\frac{5}{2}}$
$S_{5}= \frac{ \frac{32 + 243}{32} }{ \frac{5}{2} } \\ \\ S_{5}= \frac{275}{80} \\ \\ S_{5}= 3.43$

ii) If $\rm S_{5}$ =1023, r = 4, find a.

$S_{n} = \frac{a( {r}^{n} - 1)}{r - 1} \: \: \: r > 1 \\ \\ 1023 = \frac{a( {4}^{5} - 1)}{4 - 1} \\ \\ a( {4}^{5} - 1) = 1023 \times 3 \\ \\ a = \frac{1023 \times 3}{1024} \\ \\ a = 2.99$

Hope it helps you.