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

Answer: (D)

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 $\boxed{6}$.