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

Let the G.P. be \(a, ar, ar^2, ar^3, \dots, ar^{63}\)

The sum of all 64 terms in the G.P. is:
   \(S = a + ar + ar^2 + \dots + ar^{63} = \frac{a(1 - r^{64})}{1 - r}\)
The odd terms form another G.P. with first term \(a\) and common ratio \(r^2\), consisting of 32 terms. The sum of the odd terms is:
   \(S_{\text{odd}} = a + ar^2 + ar^4 + \dots + ar^{62} = \frac{a(1 - r^{64})}{1 - r^2}\)

 According to the problem, \(S = 7 \cdot S_{\text{odd}}\), so:
 \(\frac{a(1 - r^{64})}{1 - r} = 7 \cdot \frac{a(1 - r^{64})}{1 - r^2}\)
 Canceling \(a(1 - r^{64})\) from both sides (assuming \(r \neq 1\) and \(r^{64} \neq 1\)):
  \(\frac{1}{1 - r} = \frac{7}{1 - r^2}\)

Cross-multiplying gives:
 \(1 - r^2 = 7(1 - r)\)
Expanding and simplifying:
  \(r^2 - 7r + 6 = 0\)
  This is a quadratic equation in \(r\):
 \(r^2 - 7r + 6 = 0\)
 Solving this quadratic equation using the factorization method:
  \((r - 6)(r - 1) = 0\)

 Thus, \(r = 6\) or \(r = 1\). 
Since \(r = 1\) would make all terms in the G.P. identical (which does not satisfy the conditions of the problem), we conclude that:
   \(r = 6\)
  So, the common ratio of the G.P. is \(\boxed{6}\).