If in the expansion of $(1 + x)^{n}$ , the coefficient of $14^{t h}, 15^{t h}$ and $16^{t h}$ terms are in A.P. Find $n$ .

Open in App

Solution

Given: Coefficient of $14^{t h}, 15^{t h}, 16^{t h}$ term of $(1 + x)^{n}$ are in $A . P$
To find : $n$
equation: As we know,
the binomial eception of $(a + b)^{n}$ is
${a + b}^{n} =^{n} C_{0} a^{n} +^{n} C_{1} a^{n - 1} b^{1} +^{n} C_{2} a^{n - 2} b^{2} + . . . .^{n} C_{r} a^{n - r} b^{n} + . . . . .^{n} n_{0} b^{n}$
so for $(1 + x)^{n} =^{n} C_{0} (1)^{n} +^{n} C_{1} (1)^{n - 1} x^{1} +^{n} C_{2} (1)^{n - 2} x^{2} + . . . .^{n} C_{n} x^{n}$
Coefficient of $14^{t h}, 15^{t h}, 16^{t h}$ term are $^{n} C_{13},^{n} C_{14},^{n} C_{15}$
Acc. they are in $A . P$ so
$2^{n} C_{14} =^{n} C_{13} +^{n} C_{15}$ [If $a, b, c$ are in $A . P 2 b = a + c$ ]
$2 \frac{n!}{(n - 14)! 14!} = \frac{n!}{(n - 13)! 13!} + \frac{n!}{(n - 15)! 15!}$
$\frac{2}{(n - 14) (n - 15)! 14 \times 13!} = \frac{1}{(n - 13) (n - 14) (n - 15)! 13!} + \frac{1}{(n - 15)! 15 \times 14 \times 13!}$
$\frac{2}{14 (n + 14)} = \frac{1}{(n - 13) (n - 14)} + \frac{1}{210}$
$\frac{1}{n - 14} [\frac{1}{7} - \frac{1}{n - 13}] = \frac{1}{210}$
$\frac{1}{n - 14} [\frac{n - 13 - 7}{7 n - 91}] = \frac{1}{210} \Rightarrow \frac{n - 20}{7 n^{2} - 91 n - 98 n + 1274} = \frac{1}{210}$
$210 n - 4200 = 7 n^{2} - 189 n + 1274$
$7 n^{2} 189 n - 210 n + 1274 + 4200 = 0 \Rightarrow 7 n^{2} - 399 n + 5474 = 0$
$n^{2} - 57 n + 782 = 0$
$n^{2} - 34 n - 23 n + 782 = 0 n (n - 34) - 23 (n - 34) = 0$
$(n - 23) (n - 34) = 0$
$n = 23 a r 34$

Suggest Corrections

Similar questions

Q. Find

n

if the coefficient of

5^{t h}, 6^{t h}

7^{t h}

terms in the expansion of

(1 + x)^{n}

are in

A . P .

Q. If the coefficients of

5^{t h}

6^{t h}

and

7^{t h}

terms in the expansion of

{(1 + x)}^{n}

are in

A . P .

, find

n

In the binomial expansion of the (1+x)ⁿ the coefficient of fifth ,sixth and the seventh terms are in A.P . find all value of n for which this can happen

Q. The coefficients of 5th, 6th and 7th terms in the expansion of (1 + x)ⁿ are in A.P., find n.

If the coefficients of $2^{n d}$ , $3^{r d}$ and $4^{t h}$ terms in the expansion of $(1 + x)^{n}, n \in N$ are in A.P., then n =

Join BYJU'S Learning Program

If in the expansion of (1+x)n, the coefficient of 14th,15th and 16th terms are in A.P. Find n.

If in the expansion of $(1 + x)^{n}$ , the coefficient of $14^{t h}, 15^{t h}$ and $16^{t h}$ terms are in A.P. Find $n$ .