If in a G.P. of 64 terms, 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

The correct answer is 6.

Step-by-step Explanation:

Let the first term of the Geometric Progression (G.P.) be $a$ and the common ratio be $r$.
The G.P. has 64 terms, which can be represented as:
$a,ar,a{r}^2,a{r}^3,...,a{r}^{63}$

1. Sum of all 64 terms:
The sum of a G.P. of $n$ terms is given by the formula:
$S_n=\frac{a(r^n-1)}{r-1}$

For $n=64$ terms, the total sum ($S_{64}$) is:
$S_{64}=\frac{a(r^{64}-1)}{r-1}$

2. Sum of the odd terms:
The odd-positioned terms are the 1st, 3rd, 5th, ..., 63rd terms:
$a,a{r}^2,a{r}^4,...,a{r}^{62}$

This series itself forms a new G.P. where:
- The first term is $a$
- The common ratio is $r^2$
- The number of terms is $\frac{64}{2}=32$

Using the sum formula for this G.P., the sum of the odd terms ($S_{\text{odd}}$) is:
$S_{\text{odd}}=\frac{a({\left(r^2\right)}^{32}-1)}{r^2-1}=\frac{a(r^{64}-1)}{r^2-1}$

3. Relating the two sums:
According to the problem, the sum of all terms is 7 times the sum of the odd terms:
$S_{64}=7\cdot S_{\text{odd}}$

Substitute the expressions we derived:
$\frac{a(r^{64}-1)}{r-1}=7\cdot \frac{a(r^{64}-1)}{r^2-1}$

Assuming $a\ne 0$ and $r\ne \pm 1$, we can divide both sides by $a(r^{64}-1)$:
$\frac{1}{r-1}=\frac{7}{r^2-1}$

We know that $r^2-1=(r-1)(r+1)$. Substituting this in, we get:
$\frac{1}{r-1}=\frac{7}{(r-1)(r+1)}$

Multiplying both sides by $(r-1)$ simplifies the equation to:
$1=\frac{7}{r+1}$

Solving for $r$:
$r+1=7$
$r=6$

Thus, the common ratio of the G.P. is 6.