The two consecutive terms in the expansion of (3x + 2)74, whose coefficients are equal, are

Since, T3T2=7 (given)⇒  nC22xn−24−x2 nC12xn−1⋅4−x=7⇒ n−12⋅12x3=7Also,  nC2+nC1=36⇒ n(n−1)2+n=36⇒ n2+n−72=0⇒ n=8,−9n = – 9 is not possible as in Eq. (1), n – 1 should be positive.Substituting n = 8 in Eq. (1), we get23x=12=2−1⇒ 3x=−1⇒ x=−13