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Question
If the coefficients of the rth,(r+1)th & (r+2)th terms in the expansion of(1+x)14 are in A.P then the value for r is
If the coefficients of the rth,(r+1)th & (r+2)th terms in the expansion of(1+x)14 are in A.P then the value for r is
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Solution
The correct option is D both 5 and 9
Given rth,(r+1)th,(r+2)th term of (1+x)14 are in AP, we know that Tr+1=nCr(x)n is the general term of (1+x)nTr−1=14Cr−1(x)r−1Tr+1=14Cr(x)rTr+2=14Cr+1(x)r+1given,14Cr−1,14Cr,14Cr+1 are in AP214Cr=14Cr−1+14Cr+12⋅14!(14−r)!r!=14!(14−r+1)!(r−1)!+14!(14−r−1)!(r+1)!2(14−r)(13−r)!r(r−1)!=1(15−r)(14−r)(13−r)!(r−1)!+1(13−r)!(r+1)r(r−1)!2(14−r)r=1(15−r)(14−r)+1(r+1)r2(14−r)r−1(15−r)(14−r)=1(r2+r)2(15−r)(14−r)−r(14−r)(14−r)r(15−r)(14−r)=1(r2+r)2(15−r)−rr(15−r)(14−r)=1(r2+r)(30−3r)(r2+r)=r(14−r)(15−r)30r2−3r3+30r−3r2=r(210−29r+r2)30r2−3r3+30r−3r2=210r−29r2+r34r3−56r2+180r=0r(4r2−56r+180)=04r(r2−14r+45)=04r(r−9)(r−5)=0r=0,5,9r=5,9
Given rth,(r+1)th,(r+2)th term of (1+x)14 are in AP, we know that Tr+1=nCr(x)n is the general term of (1+x)n
Tr−1=14Cr−1(x)r−1Tr+1=14Cr(x)rTr+2=14Cr+1(x)r+1given,14Cr−1,14Cr,14Cr+1 are in AP
214Cr=14Cr−1+14Cr+12⋅14!(14−r)!r!=14!(14−r+1)!(r−1)!+14!(14−r−1)!(r+1)!2(14−r)(13−r)!r(r−1)!=1(15−r)(14−r)(13−r)!(r−1)!+1(13−r)!(r+1)r(r−1)!2(14−r)r=1(15−r)(14−r)+1(r+1)r2(14−r)r−1(15−r)(14−r)=1(r2+r)2(15−r)(14−r)−r(14−r)(14−r)r(15−r)(14−r)=1(r2+r)2(15−r)−rr(15−r)(14−r)=1(r2+r)(30−3r)(r2+r)=r(14−r)(15−r)30r2−3r3+30r−3r2=r(210−29r+r2)30r2−3r3+30r−3r2=210r−29r2+r34r3−56r2+180r=0r(4r2−56r+180)=04r(r2−14r+45)=04r(r−9)(r−5)=0r=0,5,9r=5,9
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