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

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Solution

The correct option is C
4

Let there be 2n terms in the given G.P. with first term a and the common ratio r.

Then, a $\frac{r^{2} - 1}{(r - 1)}$ = 5a $\frac{(r^{2} - 1)}{(r^{2} - 1)}$ ⇒ r + 1 = 5 ⇒ r = 4

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

The correct option is C 4 Let there be 2n terms in the given G.P. with first term a and the common ratio r. Then, a r2−1(r−1) = 5a (r2−1)(r2−1) ⇒ r + 1 = 5 ⇒ r = 4

The correct option is C
4

Let there be 2n terms in the given G.P. with first term a and the common ratio r.

Then, a $\frac{r^{2} - 1}{(r - 1)}$ = 5a $\frac{(r^{2} - 1)}{(r^{2} - 1)}$ ⇒ r + 1 = 5 ⇒ r = 4