Question

If an be the $n^{\text{th }}$ term of the GP of positive numbers. Let $\sum _{n=100}^{100}{a}_{2n}=\alpha$ and $\sum _{n=1}^{100}{a}_{2n-1}=\beta$ such that $\alpha \ne \beta$ then the common ratio is:

• (A) $\frac{\alpha }{\beta }$
• (B) $\frac{\beta }{\alpha }$
• (C) $\sqrt{\frac{\alpha }{\beta }}$
• (D) $\sqrt{\frac{\beta }{\alpha }}$

Answer 1

The Correct Option is A

To determine the common ratio of a geometric progression (GP), let us analyze the information provided in the problem. We are given two expressions representing the sums of terms in a GP:

1. The sum of terms with indices $\left(2n\right)$ is given by $\sum _{n=100}^{100}{a}_{2n}=\alpha$.
2. The sum of terms with indices $(2n-1)$ is given by $\sum _{n=1}^{100}{a}_{2n-1}=\beta$.

In a GP, the general term is given by:

$a_n=a{r}_n-1$,

where a is the first term and r is the common ratio. Let's express the sums:

Sum of terms for even indices:

$\sum _{n=1}^{100}{a}_{2n}=a$ ($r²$ + $r⁴$ + ... + $r²⁰⁰$) = $\alpha$

This forms a GP with first term $ar²$ and common ratio $r²$.

Sum of terms for odd indices:

$\sum _{n=1}^{100}{a}_{2n-1}=a$ (1 + $r²$ + ... + $r¹⁹⁹$) = $\beta$

This also forms a GP with first term a and common ratio $r²$.

Now, the number of terms in both sums is 100. Using the formula for the sum of a GP $S_n=\frac{\left[a_1\right(r_n-1\left)\right]}{r-1}$,

we equate and simplify the expressions for $\alpha$ and $\beta$:

1. $\alpha =\frac{ar²(r²⁰⁰-1)}{r²-1}$
2. $\beta =\frac{a(r²⁰⁰-1)}{r-1}$

Dividing $\alpha$ by $\beta$ gives us:

$\frac{\alpha }{\beta }=\frac{ar²(r²⁰⁰-1)\times (r-1)}{a(r²⁰⁰-1)\times (r²-1)}$

Canceling common terms, we get:

$\frac{\alpha }{\beta }=r²$

Thus, the common ratio r of the GP is:

Answer: $\frac{\alpha }{\beta }$.