If for some positive integer n, the coefficients of three consecutive terms in the binomial expansion of (1 + x)n + 5 are in the ratio 5 : 10 : 14, then theIargest coefficient in this expansion is

Let the three consecutive terms in the binomial expansion of (1+x)n+5 are n+5Cr−1,n+5Cr and n+5Cr+1 Now, according to the given information  n+5Cr−1:n+5Cr:n+5Cr+1=5:10:14⇒ (n+5)!(r−1)!(n−r+6)!:(n+5)!r!(n−r+5)!:(n+5)!(r+1)!(n−r+4)!=5:10:14   ⇒1(n−r+6)(n−r+5):1r(n−r+5):1(r+1)r=5:10:14  So,  rn−r+6=510 ⇒ 2r=n−r+6⇒n+6=3r  and  r+1n−r+5=57⇒ 7r+7=5n−5r+25⇒ 5n+18=12rFrom Eqs. (i) and (ii), we have n =6.So, the largest coefficient in the expansion is same as the greatest binomial coefficient =11C5 or 11C6=11!5!6!=11×10×9×8×75×4×3×2=462