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Question

Let a1,a2,a3,.. be terms of an A.P. If a1+a2+...+apa1+a2+...+aq=p2q2, p not equal to q, then a6a21 equals


A

72

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B

27

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C

1141

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D

4111

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Solution

The correct option is C

1141


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

Sn=n22a1+n-1d where a1=firstterm,d=commondifference.

We have, a1+a2+...+apa1+a2+...+aq=p2q2

p22a1+(p-1)dq22a1+(q-1)d=p2q2n=pinnumeratorandn=qindenominatorpq2a1+(p-1)d2a1+(q-1)d=p2q22a1+(p-1)d2a1+(q-1)d=pqq2a1+(p-1)d=p2a1+(q-1)d2qa1+pqd-qd=2a1p+pqd-pd2a1q-p=dq-pd=2a1

Step 2: Determine the value of a6a21

We have, an=a1+n-1d where a1=firstterm,d=commondifference

a6a21=a1+6-1da1+21-1d=a1+5×2a1a1+20×2a1d=2a1=a1+10a1a1+40a1=11a141a1a6a21=1141

Hence, option (C) is the correct answer.


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