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

The given problem is about finding the common ratio of a geometric progression (G.P.) under specific conditions. Let's solve it step by step:

Let the first term of the G.P. be \(a\) and the common ratio be \(r\). The sum of all 64 terms in the G.P. can be expressed using the formula for the sum of a geometric series: 

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

The odd terms of the G.P. form another G.P. with the first term \(a\) and the common ratio \(r^2\). Since there are 64 terms in total, and half of them are odd terms (i.e., 32 terms), their sum is given by:

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

According to the problem, the total sum of all terms is 7 times the sum of the odd terms:

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

We can simplify the equation by canceling \(a(r^{64} - 1)\) from both sides (assuming \(a \neq 0\) and \(r \neq 1\)):

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

Cross-multiplying gives us:

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

Expanding and simplifying the equation:

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

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

Factoring the quadratic equation, we get:

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

Thus, the solutions are \(r = 1\) and \(r = 6\). Since \(r = 1\) would result in a constant sequence (which doesn't satisfy the conditions unless specified otherwise), we discard it. Therefore, the common ratio is:

The correct answer is 6.

Answer 2

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 \(\boxed{6}\).