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Question

An A.P., a G.P., and a H.P. have a and b for their first two terms their (n+2)th terms will be in G.P. if b2n+2−a2n+2ab(b2n−a2n)

A
nn+1
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B
n+1n
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C
n+1n+2
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D
2n+1n+2
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Solution

The correct option is B n+1n
The (n+2)th term of A.P.
x1=a+(n+1)(ba)
The (n+2)th term of G.P.
x2=a(ba)n+1
the (n+2)th term of H.P.
x3=11a+(n+1)(1b1a)
Now ,x1, x2, x3 are in G.P. if x22=x1x3 that is
if a2(ba)2n+2=a+(n+1)(ba)1a+(n+1)(1b1a)Or b2n+2a2n=(n+1)bnab+(n+1)(ab)abb2n+2a2n+2ab(b2na2n)=n+1n

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