If the 17th and 18th terms in the expansion of

Given : $t_{17}=t_{18}$

To Find : value of a

For $(2+a)^{50}$

We have, $t_{r+1}=\left(\begin{array}{l}n \\ r\end{array}\right) A^{n-r} b^{r}$

For the $17^{\text {th }}$ term, $r=16$

$\therefore \mathrm{t}_{17}=\mathrm{t}_{16+1}$

$=\left(\begin{array}{l}50 \\ 16\end{array}\right)(2)^{50-16}(a)^{16}$

$=\left(\begin{array}{l}50 \\ 16\end{array}\right)(2)^{34}(a)^{16}$

For the $18^{\text {th }}$ term, $r=17$

$=\left(\begin{array}{l}50 \\ 17\end{array}\right)(2)^{50-17}(\mathrm{a})^{17}$

$=\left(\begin{array}{l}50 \\ 17\end{array}\right)(2)^{33}(\mathrm{a})^{17}$

As $17^{\text {th }}$ and $18^{\text {th }}$ terms are equal

$\therefore \mathrm{t}_{18}=\mathrm{t}_{17}$

$\therefore\left(\begin{array}{l}50 \\ 17\end{array}\right)(2)^{33}(a)^{17}=\left(\begin{array}{l}50 \\ 16\end{array}\right)(2)^{34}(a)^{16}$

$\therefore\left(\begin{array}{l}50 \\ 17\end{array}\right)(2)^{33}(a)^{17}=\left(\begin{array}{l}50 \\ 16\end{array}\right)(2)^{34}(a)^{16}$

$\therefore \frac{50 !}{(50-17) ! \times(17) !}(2)^{33}(a)^{17}=\frac{50 !}{(50-16) ! \times(16) !}(2)^{34}(a)^{16}$

$\left[\because\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) ! \times(\mathrm{r}) !}\right]$

$\therefore \frac{(\mathrm{a})^{17}}{(\mathrm{a})^{16}}=\frac{50 !}{(50-16) ! \times(16) !} \cdot \frac{(50-17) ! \times(17) !}{50 !} \cdot \frac{(2)^{34}}{(2)^{33}}$

$\therefore a=\frac{(50-17) \times(50-16) ! \times 17 \times(16) !}{(50-16) ! \times(16) !} .(2)$

$[\because \mathrm{n} !=\mathrm{n}(\mathrm{n}-1) !]$

$\therefore a=(50-17) \times 17 .(2)$

$\cdot a=1122$

$\underline{\text { Conclusion }}$ : value of $\mathrm{a}=1122$