JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Binomial Theorem

If some three consecutive coefficients in the binomial expansion of

(x + 1)^{n}

in powers of

x

are in the ratio

2 : 15 : 70

, then the average of these three coefficients is:

Answer: (D)

Solution

Call the three consecutive coefficients

(r - 1 n), (r n), (r + 1 n)

. The ratio of the first two,

\frac{( r - 1 n )}{( r n )} = \frac{r}{n - r + 1} = \frac{2}{15}

, gives

17 r = 2 n + 2

. The ratio of the next two,

\frac{( r n )}{( r + 1 n )} = \frac{r + 1}{n - r} = \frac{15}{70} = \frac{3}{14}

, gives

17 r = 3 n - 14

. Equating these,

2 n + 2 = 3 n - 14

, so

n = 16

and

r = 2

. The coefficients are

(1 16) = 16

(2 16) = 120

and

(3 16) = 560

, whose average is

\frac{16 + 120 + 560}{3} = \frac{696}{3} = 232

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option