(d) ${T_{r\,}} = {\,^{14}}{C_{r – 1}}\,{x^{r – 1}}\,;\,\,{T_{r + 1}}\, = \,{\,^{14}}{C_r}\,{x^r}$; ${T_{r + 2}} = {\,^{14}}{C_{r + 1}}\,{x^{r + 1}}$

By the given condition $2\,{.^{14}}{C_r} = {\,^{14}}{C_{r – 1}} + {\,^{14}}{C_{r + 1}}$ …..$(i)$

==> $2\,.\frac{{14!}}{{r!\,\,(14 – r)!}}\, = \,\frac{{14!}}{{(r – 1)!\,\,(15 – r)!}} + \frac{{14!}}{{(r + 1)!\,(13 – r)!}}$

==> $\frac{2}{{r.(r – 1)!.(14 – r).(13 – r)!}}$=$\frac{1}{{(r – 1)!.(15 – r).(14 – r).(13 – r)!}}$

+$\frac{1}{{(r + 1)\,r(r – 1)\,!\,(13 – r)\,!}}$

==> $\frac{2}{{r(14 – r)}} = \frac{1}{{(15 – r)(14 – r)}} + \frac{1}{{(r + 1)r}}$

==> $\frac{1}{{r(14 – r)}} – \frac{1}{{(15 – r)(14 – r)}} = \frac{1}{{(r + 1)r}} – \frac{1}{{r(14 – r)}}$

==> $\frac{{(15 – r) – r}}{{r(15 – r)(14 – r)}} = \frac{{(14 – r) – (r + 1)}}{{(r + 1)r(14 – r)}}$

==> $\frac{{15 – 2r}}{{15 – r}} = \frac{{13 – 2r}}{{r + 1}}$

==> $15r + 15 – 2{r^2} – 2r = 195 – 30r – 13r + 2{r^2}$

==> $4{r^2} – 56r + 180 = 0\,\,$

$\Rightarrow \,{r^2} – 14r + 45 = 0$

==> $(r – 5)(r – 9) = 0 $

$\Rightarrow r = 5,\,9$

But $5$ is not given.

Hence $r = 9.$ 

$Trick :$ Put the value of $r$ from options in equation $(i)$, only $(d)$ satisfy it.