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 the largest coefficient in the expansion is :

Correct answer: A. 462 Let three consecutive terms are T _ r , T _ r + 1 , T _ r + 2 So, (Cr - 1 n + 5)/(Cr n + 5) = (5)/(10) and (Cr n + 5)/(Cr + 1 n + 5) = (10)/(14) Using property (Cr n)/(Cr - 1 n) = (n - r + 1)/(r) , we have ((n + 5) - r + 1)/(r) = 2 and ((n + 5) - (r + 1) + 1)/(r + 1) = (7)/(5) ⇒ n - 3r + 6 = 0 and 5n - 12r + 18 = 0 Solving both equations, n = 6, r = 4 Hence, the greatest coefficient will be of middle term, which is = C5 n + 5 = C5 11 = 462

JEE Main 20204 SeptemberEvening Shift

Exam: JEE Main
Year: 2020
Date: 4 September
Shift: Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2020 · 4 September · Evening Shift · Mathematics · Binomial Theorem

If for some positive integer n , the coefficients of three...

If for some positive integer

n

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

{\left(1 + x\right)}^{n + 5}

are in the ratio

5: 10: 14

, then the largest coefficient in the expansion is :

Options

A $462$
B $330$
C $792$
D $252$

Correct answer

A. $462$

Step-by-step solution

Let three consecutive terms are ${T}_{r}, {T}_{r + 1}, {T}_{r + 2}$
So, $\dfrac{Cr - 1 n + 5}{Cr n + 5} = \dfrac{5}{10}$ and $\dfrac{Cr n + 5}{Cr + 1 n + 5} = \dfrac{10}{14}$

Using property $\dfrac{Cr n}{Cr - 1 n} = \dfrac{n - r + 1}{r}$ , we have

$\dfrac{\left(n + 5\right) - r + 1}{r} = 2$ and $\dfrac{\left(n + 5\right) - \left(r + 1\right) + 1}{r + 1} = \dfrac{7}{5}$

$⇒ n - 3r + 6 = 0$ and $5n - 12r + 18 = 0$

Solving both equations, $n = 6, r = 4$
Hence, the greatest coefficient will be of middle term, which is $= C5 n + 5 = C5 11 = 462$

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