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Question

If a,b,c are in GP and a,p,q are in AP such that 2a,b+p,c+q are in GP then the common difference of AP is

A
2a
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B
(2+1)(ab)
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C
2(a+b)
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D
(21)(ba)
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Solution

The correct options are
B (2+1)(ab)
D (21)(ba)
Let common difference of the A.P is d.
Given:
b2=ac .......(1)
2p=a+q .......(2)
(b+p)2=(c+q)2a .......(3)
(3)-(1) we get, p(2b+p)=ac+2aq
b2(2p)b+2aqp2=0 from (1)
b2(a+q)+8aq(a+q)24=0
b=a+q±a2+q2+2aq+a2+q2+2aq8aq2
2b=a+q±2a2+2q24aq
=a+q±2(qa)
=a(12)+q(1±2)
=a(12)+(a+2d)(1±2)
2b=2a+2d(1±2)
d=ba1±2
or d=(21)(ba),(2+1)(ab)
Hence, Options B and D are correct.

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