A G.P. consists of an even number of terms and let the terms be a, ar, $ar^2, ar^3$ , ... , $ar^{2n-1}$ i.e., it contains 2n terms. According to question, Sum of all terms = 5 (sum of odd places terms) ⇒ a + $ar+ar^2$ + ... + $ar^{2n-1}$ = $5(a+ar^2+\dots+ar^{2n-2})$ ⇒ $\frac{a(1-r^{2n})}{1-r}$ = $5\left[\frac{a[1-(r^2)^n]}{1-r^2}\right]$ ⇒ $\frac{a(1-r^{2n})}{1-r^2}$ ⇒ $\frac{1}{1-r}$ = $\frac{5}{(1-r)(1+r)}$ ⇒ r + 1 = 5 ⇒ r = 4