Explanation

(d)
$T_{r} = \ ^{14}C_{r - 1}x^{r - 1};\ T_{r + 1} = \ ^{14}C_{r}x^{r};\ T_{r + 2} = \ ^{14}C_{r + 1}x^{r + 1}$
Since, these terms are in AP
$\therefore\ {2T}_{r + 1} = T_{r} + T_{r + 2}$
$\Rightarrow 2\ ^{14}C_{r} = \ ^{14}C_{r - 1} + \ ^{14}C_{r + 1}$ …(i)
$\Rightarrow 2 \bullet \frac{14!}{r!(14 - r)!} = \frac{14!}{(r - 1)!(15 - r)!}$
$+ \frac{14!}{(r + 1)!(13 - r)!}$
$\Rightarrow \frac{2}{r \bullet (r - 1)!(14 - r) \bullet (13 - r)!}$
$= \frac{1}{(r - 1)! \bullet (15 - r) \bullet (14 - r) \bullet (13 - r)!}$
$+ \frac{1}{(r + 1)r(r - 1)!(13 - r)!}$
$\Rightarrow \ \frac{2}{r(14 - r)} = \frac{1}{(15 - r)(14 - r)} + \frac{1}{(r + 1)r}$
$\Rightarrow \ \frac{1}{r(14 - r)} = \frac{1}{(15 - r)(14 - r)}$
$+ \frac{1}{(r + 1)r} - \frac{1}{r(14 - r)}$
$\Rightarrow \ \frac{(15 - r) - r}{r(15 - r)(14 - r)} = \frac{(14 - r) - (r + 1)}{(r + 1)r(14 - r)}$
$\Rightarrow \ (15 - r) - r = (13 - 2r)(15 - r)$
$\Rightarrow \ 15r + 15 - 2r^{2} - 2r = 195 - 30r - 13r + 2r^{2}$
$\Rightarrow \ 4r^{2} - 56r + 180 = 0$
$\Rightarrow \ r^{2} - 14r + 45 = 0$
$\Rightarrow \ (r - 5)(r - 9) = 0\ \Rightarrow \ r = 5,9$
But 5 is not given.
Hence, $r = 9$