A GP consists of an even number of terms.

Question:

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

 

Solution:

Let the terms of the G.P. be $a, a r, a r^{2}, a r^{3}, \ldots, a r^{n-2}, a r^{n-1}$

Sum of a G.P. series is represented by the formula, $\mathrm{S}_{\mathrm{n}}=\mathrm{a} \frac{\mathrm{r}^{\mathrm{n}}-1}{\mathrm{r}-1}$

when r≠1. ‘Sn’ represents the sum of the G.P. series upto nth terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Thus, the sum of this G.P. series is $S_{n}=a \frac{r^{n}-1}{r-1}$

The odd terms of this series are $a, a r^{2}, a r^{4}, \ldots, a r^{n-2}$

\{Since the number of terms of the G.P. series is even; the $2^{\text {nd }}$ last term will be an odd term. $\}$

Here,

No. of terms will be $\frac{\mathrm{n}}{2}$ as we are splitting up the n terms into 2 equal parts of odd and even terms. {since the no. of terms is even, we have 2 equal groups of odd and even terms }

Sum of the odd terms ⇒

$\mathrm{S}_{\mathrm{n}}=\mathrm{a} \times \frac{\mathrm{r}^{2\left(\frac{\mathrm{n}}{2}\right)}-1}{\mathrm{r}^{2}-1}$

$\Rightarrow \mathrm{S}_{\mathrm{n}}=\mathrm{a} \times \frac{\mathrm{r}^{\mathrm{n}}-1}{(\mathrm{r}-1)(\mathrm{r}+1)}$

By the problem,

$a \frac{r^{n}-1}{r-1}=5 \times a \times \frac{r^{n}-1}{(r-1)(r+1)}$

$\Rightarrow 1=\frac{5}{(r+1)}$

$\Rightarrow r+1=5$

$\Rightarrow \therefore r=4$

Thus, the common ratio (r) = 4

 

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