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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

$\frac{7}{2}$

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B

$\frac{2}{7}$

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C

$\frac{11}{41}$

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D

$\frac{41}{11}$

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Solution

## 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${S}_{n}=\frac{n}{2}\left[2{a}_{1}+\left(n-1\right)d\right]$ where ${a}_{1}=\text{firstterm},d=\text{commondifference}$.We have, $\frac{{a}_{1}+{a}_{2}+...+{a}_{p}}{{a}_{1}+{a}_{2}+...+{a}_{q}}=\frac{{p}^{2}}{{q}^{2}}$$â‡’\frac{\frac{p}{2}\left[2{a}_{1}+\left(p-1\right)d\right]}{\frac{q}{2}\left[2{a}_{1}+\left(q-1\right)d\right]}=\frac{{p}^{2}}{{q}^{2}}\left[âˆµn=p\text{innumeratorand}n=q\text{indenominator}\right]\phantom{\rule{0ex}{0ex}}â‡’\frac{p}{q}\frac{\left[2{a}_{1}+\left(p-1\right)d\right]}{\left[2{a}_{1}+\left(q-1\right)d\right]}=\frac{{p}^{2}}{{q}^{2}}\phantom{\rule{0ex}{0ex}}â‡’\frac{\left[2{a}_{1}+\left(p-1\right)d\right]}{\left[2{a}_{1}+\left(q-1\right)d\right]}=\frac{p}{q}\phantom{\rule{0ex}{0ex}}â‡’q\left[2{a}_{1}+\left(p-1\right)d\right]=p\left[2{a}_{1}+\left(q-1\right)d\right]\phantom{\rule{0ex}{0ex}}â‡’2q{a}_{1}+\overline{)pqd}-qd=2{a}_{1}p+\overline{)pqd}-pd\phantom{\rule{0ex}{0ex}}â‡’2{a}_{1}\overline{)\left(q-p\right)}=d\overline{)\left(q-p\right)}\phantom{\rule{0ex}{0ex}}â‡’d=2{a}_{1}$Step 2: Determine the value of $\frac{{a}_{6}}{{a}_{21}}$We have, ${a}_{n}={a}_{1}+\left(n-1\right)d$ where ${a}_{1}=\text{firstterm},d=\text{commondifference}$ $âˆ´\frac{{a}_{6}}{{a}_{21}}=\frac{{a}_{1}+\left(6-1\right)d}{{a}_{1}+\left(21-1\right)d}\phantom{\rule{0ex}{0ex}}=\frac{{a}_{1}+\left(5Ã—2{a}_{1}\right)}{{a}_{1}+20Ã—2{a}_{1}}\left[âˆµd=2{a}_{1}\right]\phantom{\rule{0ex}{0ex}}=\frac{{a}_{1}+10{a}_{1}}{{a}_{1}+40{a}_{1}}\phantom{\rule{0ex}{0ex}}=\frac{11{a}_{1}}{41{a}_{1}}\phantom{\rule{0ex}{0ex}}\frac{{a}_{6}}{{a}_{21}}=\frac{11}{41}$Hence, option (C) is the correct answer.

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