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Question: Let a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ...... be terms of an A.P. If \(\frac{a_{1} + a_{2}...

Let a1, a2, a3, ...... be terms of an A.P. If a1+a2+....+apa1+a2+.....+aq\frac{a_{1} + a_{2} + .... + a_{p}}{a_{1} + a_{2} + ..... + a_{q}} = p2q2\frac{p^{2}}{q^{2}}, p ¹ q, then a6a21\frac{a_{6}}{a_{21}} equals-

A

41/11

B

7/2

C

2/7

D

11/41

Answer

11/41

Explanation

Solution

a1+a2+......+apa1+a2+......+aq\frac{a_{1} + a_{2} + ...... + a_{p}}{a_{1} + a_{2} + ...... + a_{q}} = p2q2\frac{p^{2}}{q^{2}}

a1+a2+......+a6a1+a2+......+aq\frac{a_{1} + a_{2} + ...... + a_{6}}{a_{1} + a_{2} + ...... + a_{q}} = 62q2\frac{6^{2}}{q^{2}}

a1+a2+......+a5a1+a2+......+aq\frac{a_{1} + a_{2} + ...... + a_{5}}{a_{1} + a_{2} + ...... + a_{q}} = 52q2\frac{5^{2}}{q^{2}}

on subtracting

a6a1+....+qq=62−52q2\frac{a_{6}}{a_{1} + .... + q_{q}} = \frac{6^{2} - 5^{2}}{q^{2}}

Similarly a21a1+....+qq=(21)2−(20)2q2\frac{a_{21}}{a_{1} + .... + q_{q}} = \frac{(21)^{2} - (20)^{2}}{q^{2}}