If coefficients of 2nd ,3rd  and 4th  term in the expansion of (1+x)2n are in A.P., then

By the given condition, C1   2n,C2   2n,C3   2n are in A.P.⇒22nC2=C1   2n+C3   2n ⇒2·2n(2n-1)1.2=2n1+2n(2n-1)2n-21.2.3⇒2n-1=1+(2n-1)(n-1)3⇒3+2n2-3n+1=6n-3⇒2n2-9n+7=0