A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then the common ratio is

Solution :
Let the G.P. be $$a,ar,a{r^2},.....$$
$$S = a + ar + a{r^2} + ..... +\, {\text{to }}2n{\text{ term}} = \frac{{a\left( {{r^{2n}} - 1} \right)}}{{r - 1}}$$
The series formed by taking term occupying odd places is $${S_1} = a + a{r^2} + a{r^4} + .....\,{\text{to }}n{\text{ terms}}$$
$$\eqalign{ & {S_1} = \frac{{a\left[ {{{\left( {{r^2}} \right)}^n} - 1} \right]}}{{{r^2} - 1}} \cr & \Rightarrow {S_1} = \frac{{a\left( {{r^{2n}} - 1} \right)}}{{{r^2} - 1}} \cr & {\text{Now, }}S = 5{S_1}{\text{ or }}\frac{{a\left( {{r^{2n}} - 1} \right)}}{{r - 1}} = 5\frac{{a\left( {{r^{2n}} - 1} \right)}}{{{r^2} - 1}} \cr & \Rightarrow 1 = \frac{5}{{r + 1}} \cr & \Rightarrow r + 1 = 5 \cr & \therefore r = 4 \cr} $$