Question

Let \(a_1,a_2,a_3,\ldots\) be the terms of an A.P. If \[ \frac{a_1+a_2+\cdots+a_p}{a_1+a_2+\cdots+a_q}=\frac{p^2}{q^2}\quad (p\neq q), \] then \(\dfrac{a_6}{a_{21}}=\)

• A. \(\dfrac{7}{2}\)
• B. \(\dfrac{2}{7}\)
• C. \(\dfrac{11}{41}\)
• D. \(\dfrac{41}{11}\)

Hint:

If the ratio of sums of an A.P. depends on \(n^2\), then:
Compare coefficients of \(n^2\) and \(n\) in \(S_n\)
This often gives a condition relating \(a\) and \(d\)

Answer 1

The Correct Option is C

Step 1: Let the A.P. have first term \(a\) and common difference \(d\). Sum of first \(n\) terms: \[ S_n=\frac{n}{2}[2a+(n-1)d] \]
Step 2: Given: \[ \frac{S_p}{S_q}=\frac{p^2}{q^2} \] Substitute the formula for sums: \[ \frac{p[2a+(p-1)d]}{q[2a+(q-1)d]}=\frac{p^2}{q^2} \] Cancel \(\frac{p}{q}\): \[ \frac{2a+(p-1)d}{2a+(q-1)d}=\frac{p}{q} \]
Step 3: Cross-multiply: \[ q[2a+(p-1)d]=p[2a+(q-1)d] \] \[ 2aq+qd(p-1)=2ap+pd(q-1) \] Simplifying: \[ 2a(q-p)=d(q-p) \] Since \(p\neq q\), \[ 2a=d \Rightarrow a=\frac{d}{2} \]
Step 4: General term of the A.P.: \[ a_n=a+(n-1)d=\frac{d}{2}+(n-1)d=\frac{(2n-1)d}{2} \]
Step 5: Compute the required ratio: \[ \frac{a_6}{a_{21}}=\frac{\frac{(2\cdot6-1)d}{2}}{\frac{(2\cdot21-1)d}{2}} =\frac{11}{41} \]