Question

Let $a_{1,}{a}_2,a_3,..$ be terms of an A.P. If $\frac{a_1+a_2+...+a_p}{a_1+a_2+...+a_q}=\frac{p^2}{q^2}$, $p$ not equal to $q$, then $\frac{a_6}{a_{21}}$ equals

• A. $\frac{7}{2}$
• B. $\frac{2}{7}$
• C. $\frac{11}{41}$
• D. $\frac{41}{11}$

Answer 1

The correct option is C

$\frac{11}{41}$

Step 1: Determine the common difference $d$

An arithmetic progression (AP) is a progression in which the difference between two consecutive terms is constant.

Sum of $n$ terms of AP, is given by

$S_n=\frac{n}{2}\left[2a_1+\left(n-1\right)d\right]$ where $a_1=\text{first term},d=\text{common difference}$.

We have, $\frac{a_1+a_2+...+a_p}{a_1+a_2+...+a_q}=\frac{p^2}{q^2}$

$$\begin{gathered} \Rightarrow \frac{{\displaystyle \frac{p}{2}}\left[2a_1+(p-1)d\right]}{{\displaystyle \frac{q}{2}}\left[2a_1+(q-1)d\right]}=\frac{p^2}{q^2}\left[\because n=p\text{in numerator and}n=q\text{in denominator}\right] \\ \Rightarrow \frac{p}{q}\frac{{\displaystyle \left[2a_1+(p-1)d\right]}}{{\displaystyle \left[2a_1+(q-1)d\right]}}=\frac{p^2}{q^2} \\ \Rightarrow \frac{{\displaystyle \left[2a_1+(p-1)d\right]}}{{\displaystyle \left[2a_1+(q-1)d\right]}}=\frac{p}{q} \\ \Rightarrow q\left[2a_1+(p-1)d\right]=p\left[2a_1+(q-1)d\right] \\ \Rightarrow 2q{a}_1+\cancel{pqd}-qd=2a_{1}p+\cancel{pqd}-pd \\ \Rightarrow 2a_1\cancel{\left(q-p\right)}=d\cancel{\left(q-p\right)} \\ \Rightarrow d=2a_1 \end{gathered}$$

Step 2: Determine the value of $\frac{a_6}{a_{21}}$

We have, $a_n=a_1+\left(n-1\right)d$ where $a_1=\text{first term},d=\text{common difference}$

$$\begin{gathered} \therefore \frac{a_6}{a_{21}}=\frac{a_1+\left(6-1\right)d}{a_1+\left(21-1\right)d} \\ =\frac{a_1+\left(5\times 2a_1\right)}{a_1+20\times 2a_1}\left[\because d=2a_1\right] \\ =\frac{a_1+10a_1}{a_1+40a_1} \\ =\frac{11a_1}{41a_1} \\ \frac{a_6}{a_{21}}=\frac{11}{41} \end{gathered}$$

Hence, option (C) is the correct answer.