if the coefficients of three consecutive terms in the expansion of (1 + x) ^ n are the ratio 1: 5: 20then the coefficient of the fourth term is

Correct answer: D. 3654 Given, Cr - 1 n : Cr n : Cr + 1 n = 1: 5: 20 Now using the formula (Cr n)/(Cr - 1 n) = (n - r + 1)/(r) we get, . (Cr n)/(Cr - 1 n) = (n - r + 1)/(r) = (5)/(1) ⇒ n - 6r + 1 = 0 . . . . . . . . . (1) And (Cr + 1 n)/(Cr n) = (n - r)/(r + 1) = (20)/(5) ⇒ n - 5r - 4 = 0 . . . . . . . . . . (2) Solving above equations we get, n = 29 and r = 5 So, the coefficient of forth term will be C3 n = C3 29 = (29 × 28 × 27)/(3 × 2 × 1) = 3654

JEE Main 20238 AprilMorning Shift

Exam: JEE Main
Year: 2023
Date: 8 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2023 · 8 April · Morning Shift · Mathematics · Binomial Theorem

if the coefficients of three consecutive terms in the expansion of...

if the coefficients of three consecutive terms in the expansion of

{\left(1 + x\right)}^{n} are the ratio

1: 5: 20then the coefficient of the fourth term is

Options

A $2436$
B $5481$
C $1827$
D $3654$

Correct answer

D. $3654$

Step-by-step solution

Given,
$Cr - 1 n : Cr n : Cr + 1 n = 1: 5: 20$
Now using the formula $\dfrac{Cr n}{Cr - 1 n} = \dfrac{n - r + 1}{r}$ we get,
. $\dfrac{Cr n}{Cr - 1 n} = \dfrac{n - r + 1}{r} = \dfrac{5}{1}$
$⇒ n - 6r + 1 = 0 . . . . . . . . . \left(1\right)$
And $\dfrac{Cr + 1 n}{Cr n} = \dfrac{n - r}{r + 1} = \dfrac{20}{5}$
$⇒ n - 5r - 4 = 0 . . . . . . . . . . \left(2\right)$
Solving above equations we get,
$n = 29 \text{ and } r = 5$
So, the coefficient of forth term will be $C3 n = C3 29 = \dfrac{29 \times 28 \times 27}{3 \times 2 \times 1} = 3654$

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