Question

If in a G.P. of64terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P., then the common ratio of the G.P. is equal to:

• A. 7
• B. 4
• C. 5
• D. 6

Answer 1

The Correct Option is D

This problem requires determining the common ratio of a geometric progression (G.P.) given specific constraints. The solution proceeds as follows:

Let the initial term of the G.P. be denoted by \(a\) and its common ratio by \(r\). The sum of the 64 terms in the G.P. is calculated using the geometric series sum formula:

\(S_{64} = \frac{a(r^{64} - 1)}{r - 1}\)

The odd-numbered terms of the G.P. constitute a separate G.P. with the first term \(a\) and a common ratio of \(r^2\). With 64 total terms, there are 32 odd terms, and their sum is:

\(S_{32} = \frac{a(r^{64} - 1)}{r^2 - 1}\)

The problem states that the total sum of all terms is seven times the sum of the odd terms:

\(\frac{a(r^{64} - 1)}{r - 1} = 7 \times \frac{a(r^{64} - 1)}{r^2 - 1}\)

Assuming \(a eq 0\) and \(r eq 1\), we can simplify the equation by canceling \(a(r^{64} - 1)\) from both sides:

\(\frac{1}{r - 1} = \frac{7}{r^2 - 1}\)

Cross-multiplication yields:

\(r^2 - 1 = 7(r - 1)\)

Expanding and simplifying the equation results in:

\(r^2 - 1 = 7r - 7\)

\(r^2 - 7r + 6 = 0\)

Factoring the quadratic equation gives:

\((r - 1)(r - 6) = 0\)

The potential solutions are \(r = 1\) and \(r = 6\). The case \(r = 1\) implies a constant sequence, which generally does not satisfy the given conditions unless explicitly stated. Therefore, \(r = 1\) is excluded. Consequently, the common ratio is:

The correct answer is 6.