In an arithmetic progression of $16$ distinct terms with $a_1=16$, the sum is equal to square of the last term. The common difference of the A.P. is

We are given
$\begin{array}{l} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{16}{2}\left[a_1+a_{16}\right]=a_{16}^2\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} a_{16}^2-8a_{16}-\left(8\right)\left(16\right)=0\\ \Rightarrow \hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(a_{16}-16\right)\left(a_{16}+8\right)=0\end{array}$

As    $a_{16}\ne 16=a_1,a_{16}=-8$
Now, $a_{16}-a_1=-24\Rightarrow 15d=-24$
$\Rightarrow d=-8/5$