If a>0  and coefficient of x2, x3, x4 in the expansion of 1+xa6  are in A.P., then a equals

Coefficient of xr in the expansion of (1+x/a)6=6Cr(1/a)r According to given condition   6C2(1/a)2,6C3(1/a)3,6C4(1/a)4are in A.P.∴ 2 6C31a3=6C21a2+6C41a4⇒ 2(20)a=15a2+1⇒ 3a2−8a+3=0⇒ a=(4+7)3 as a>0.