Sequences & Series Question - None

Question

Let $a_1, a_2, a_3, \ldots$ be terms of an A.P. If $\dfrac{a_1 + a_2 + \cdots + a_p}{a_1 + a_2 + \cdots + a_q} = \dfrac{p^2}{q^2}$, $p \neq q$, then $\dfrac{a_6}{a_{21}}$ equals

Accepted Answer

Step 1: For an A.P., $S_n = \frac{n}{2}(2a_1 + (n-1)d)$. Given: $\frac{S_p}{S_q} = \frac{p^2}{q^2}$

Step 2: $\frac{\frac{p}{2}(2a_1 + (p-1)d)}{\frac{q}{2}(2a_1 + (q-1)d)} = \frac{p^2}{q^2}$

Step 3: Simplify: $\frac{p(2a_1 + (p-1)d)}{q(2a_1 + (q-1)d)} = \frac{p^2}{q^2}$

$\Rightarrow q(2a_1 + (p-1)d) = p(2a_1 + (q-1)d)$

Step 4: $2qa_1 + q(p-1)d = 2pa_1 + p(q-1)d$

$2(q-p)a_1 = [p(q-1) - q(p-1)]d = (pq - p - pq + q)d = (q-p)d$

$\Rightarrow 2a_1 = d$ (since $p eq q$)

Step 5: Now $a_n = a_1 + (n-1)d = a_1 + (n-1)(2a_1) = a_1(2n-1)$

Step 6: $a_6 = a_1(2 \cdot 6 - 1) = 11a_1$ and $a_{21} = a_1(2 \cdot 21 - 1) = 41a_1$

$ herefore \frac{a_6}{a_{21}} = \frac{11a_1}{41a_1} = \frac{11}{41}$

Question:

Step-by-Step Solution

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