If the coefficients of rth,(r+1)th and (r+2)th terms in the binomial expansion of 1+ym are in A.P., then m and r satisfy the equation:
m2-m(4r-1)+4r2+2=0
m2-m(4r-1)+4r2-2=0
m2-m(4r+1)+4r2-2=0
m2-n(4r-1)+4r2-2=0
Explanation for the correct option:
Given that, rth,(r+1)th and (r+2)thterm in the expansion of 1+ym are in A.P.
Since, Cr-1m,Crm,Cr+1m are in A.P.2mCr=Cr-1m+Cr+1m2=Cr-1mCrm+Cr+1mCrm2=m!r-1!m-r+1!m!r!m-r!+m!r+1!m-r-1!m!r!m-r!2=rm-r+1+m-rr+1∵CrnCr-1n=n-r+1r2=rr+1+m-rm-r+1m-r+1r+12=r2+r+m2-mr+m-mr+r2-rmr+m-r2-r+r+12=2r2+m2-2mr+mmr-r2+m+1m2-m(4r+1)+4r2-2=0Hence, option (C) is correct.