Q.

If the coefficients of P th, ( p + 1)th and (p + 2)th terms in the expansion of (1 + x)n are in AP, then

Moderate

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n2−2np+4p2=0

n2−n(4p+1)+4p2−2=0

n2−n(4p+1)+4p2=0

None of these above

answer is 2.

(Detailed Solution Below)

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Detailed Solution

coefficients of pth , (p + 1)th and (p + z)th terms in the expansion (1 + x)" are nCp−1, nCp′ nCp+1, respectively.Since, these are in AP.∴ 2nCp=nCp−1+nCp+1⇒2n!(n−p)!p!=n!(n−p+1)!(p−1)!+n!(n−p−1)!(p+1)! ⇒2(n−p)!p!=p(n−p+1)(n−p)!p!+n−p(n−p)!(p+1)p! ⇒ 21=p(n−p+1)+n−pp+1⇒ n2−n(4p+1)+4p2−2=0

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