COMPASSLearning

Talk to us Register Login

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Binomial Theorem

If the coefficients of

x^{4}

,

x^{5}

and

x^{6}

in the expansion of

(1 + x)^{n}

are in the arithmetic progression, then the maximum value of

n

is

(A)7
(B)28
(C)14✓ Correct
(D)21

Answer: (C)

Solution

(4 n), (5 n), (6 n)

in AP:

2 (5 n) = (4 n) + (6 n)

.

2 = \frac{( 4 n )}{( 5 n )} + \frac{( 6 n )}{( 5 n )} = \frac{5}{n - 4} + \frac{n - 5}{6}

12 (n - 4) = 30 + (n - 5) (n - 4) = 30 + n^{2} - 9 n + 20 \Rightarrow n^{2} - 21 n + 98 = 0 \Rightarrow (n - 14) (n - 7) = 0

Maximum

n = 14

.

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Binomial Theorem →

More JEE Main questions on Binomial Theorem

If the coefficient of $x$ in the expansion of $(a x^{2} + b x + c) (1 - 2 x)^{26}$ is -56 and the coefficients of $x^{2}$ and $x^{3}$ are both zero, then $a + b + c$ is equal to
JEE Main 2026View solution →
If …
JEE Main 2026View solution →
The coefficient of $x^{48}$ in $(1 + x) + 2 (1 + x)^{2} + 3 (1 + x)^{3} + \dots + 100 (1 + x)^{100}$ is equal to:
JEE Main 2026View solution →
Let $C_{r}$ denote the coefficient of $x^{r}$ in the binomial expansion of $(1 + x)^{n}, n \in N, 0 \leq r \leq n$ .

If …
JEE Main 2026View solution →
The value of $\frac{^{100} C _{50}}{51} + \frac{^{100} C _{51}}{52} + \dots . + \frac{^{100} C _{100}}{101}$ is :
JEE Main 2026View solution →
The sum of all possible values of $n \in N$ , so that the coefficients of $x, x^{2}$ and $x^{3}$ in the expansion of $(1 + x^{2})^{2} (1 + x)^{n}$ , are in arithmetic progression is:
JEE Main 2026View solution →