Let the coefficients of three consecutive terms in the binomial expansion of $(1+2 \mathrm{x})^{\mathrm{n}}$ be in the ratio $2: 5: 8$. Then the coefficient of the term, which is in the middle of these three terms, is ____.

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Binomial Theorem

Let the coefficients of three consecutive terms in the binomial expansion of

(1 + 2 x)^{n}

be in the ratio

2 : 5 : 8

. Then the coefficient of the term, which is in the middle of these three terms, is ____.

Answer: 1120

Let the three consecutive terms carry coefficients

(r - 1 n) 2^{r - 1}

(r n) 2^{r}

(r + 1 n) 2^{r + 1}

. The first ratio

\frac{( r - 1 n ) 2 ^{r - 1}}{( r n ) 2 ^{r}} = \frac{r}{2 ( n - r + 1 )} = \frac{2}{5}

gives

9 r = 4 (n + 1)

, and the second

\frac{( r n ) 2 ^{r}}{( r + 1 n ) 2 ^{r + 1}} = \frac{r + 1}{2 ( n - r )} = \frac{5}{8}

gives

9 r = 5 n - 4

. Solving the pair gives

n = 8

and

r = 4

, so the middle of the three coefficients is

(4 8) 2^{4} = 70 \times 16 = 1120

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Integer answer