Explanation

(c)
Coefficients of $T_{5} = \ ^{n}C_{4},\ T_{6} = \ ^{n}C_{5}$ and $T_{7} = \ ^{n}C_{6}$
According to the given condition,
$2\ ^{n}C_{5} = \ ^{n}C_{4} + \ ^{n}C_{6}$
$\Rightarrow \ 2\left\lbrack \frac{n!}{(n - 5)!5!} \right\rbrack = \left\lbrack \frac{n!}{(n - 4)!4!} + \frac{n!}{(n - 6)!6!} \right\rbrack$
$\Rightarrow \ 2\left\lbrack \frac{6}{(n - 5)} \right\rbrack = \left\lbrack \frac{5.6}{(n - 4)(n - 5)} + 1 \right\rbrack$
$\Rightarrow \ \frac{12}{(n - 5)} = \frac{30 + n^{2} - 9n + 20}{(n - 4)(n - 5)}$
$\Rightarrow \ n^{2} - 21n + 98 = 0$
$\Rightarrow \ (n - 7)(n - 14) = 0$
$\Rightarrow \ n = 7\ or\ 14$