The number of the terms of a geometric progression is even. The sum of all terms of the progression is thrice as large as the sum of its odd terms. Find the common ratio of the progression.

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Solution

From the given question, it is clear that:

$(A_{1} + A_{2} + . . . . . . . . . + A_{2 n}) = 3 \times (A_{1} + A_{2} + . . . . . . . . . + A_{2 n - 1})$

$∴ (A_{1} + A_{2} + . . . . . . . . . + A_{2 n}) - 3 \times (A_{1} + A_{2} + . . . . . . . . . + A_{2 n - 1}) = 0$

$(A_{2} + A_{4} + . . . . . . . . . + A_{2 n}) = 2 \times (A_{1} + A_{2} + . . . . . . . . . + A_{2 n - 1})$

Now if we assume that G.P. is in form $x, x d, x d^{2}, . . . . . . . .$

where x is the first terms and d is the common ratio.

Then,

$x d \frac{d^{n} - 1}{d - 1} = 2 x \frac{d^{n} - 1}{d - 1}$

$2 x = x d \Rightarrow d = 2$

Therefore the common ratio of G.P. is 2

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