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Q.
In the binomial expansion of $(1 + x)^n$, where $n$ is a natural number, the co-efficient of 5th, 6th and 7th terms are in A.P., then $n$ is equal to
Binomial Theorem
Solution:
Since co-eff. of $5th$, $6th$, $7th$ terms are in $A.P$.
$\therefore \,{}^{n}c_{4}+\,{}^{n}c_{6} = 2\cdot^{n}c_{5}$
$\Rightarrow \frac{n\,!}{n-4\,!\,4\,!}+\frac{n\,!}{n-6\,!\,6\,!}$
$= \frac{n\,!}{2n-5\,!\,5\,!}$
$\Rightarrow 6\cdot5 +\left(n-4\right)\,\left(n-5\right) =2\left(n-4\right)\cdot6$
$\Rightarrow 30 + n^{2} - 9n + 60 = 12n - 48$
$\Rightarrow n^{2} - 21n + 98 = 0$
$\Rightarrow n = 7$, $14$