Question

For the following GP.s, find $\rm S_{n}$
i) 2, -2, 2, -2, ...
ii) 0.03, 0.06, 0.12, 0.24, ...

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) 2, -2, 2, -2, ...

here a= 2

r = -1

Sum of n terms is given by second formula

$S_{n} = \frac{a( 1 - {r}^{n} )}{1 - r } \: \: \: r < 1 \\ \\ S_{n} = \frac{2( 1 - {( - 1)}^{n} )}{1 - ( - 1) } \: \: \: r < 1 \\ \\ S_{n}= \frac{2(1 - ( { - 1)}^{n}) }{2} \\ \\ S_{n}= (1 - ( { - 1)}^{n})$
ii) 0.03, 0.06, 0.12, 0.24, ...

a = 0.03

r= 2

since r > 1

$S_{n} = \frac{a( {r}^{n} - 1)}{r - 1} \: \: \: r > 1 \\ \\ S_{n} = \frac{0.03( {2}^{n} - 1)}{2- 1} \: \: \: \\ \\S_{n}= 0.03( {2}^{n} - 1)$

Hope it helps you.