Let sum and product of the mean and variance of a binomial distribution are 82.5 and1350 respectively. Then the number of trials in the binomial distribution is :

Let, mean $=\mathrm{m}=\mathrm{np}$
& variance $=\mathrm{v}=\mathrm{npq},\mathrm{p}+\mathrm{q}=1$
Sum $=\mathrm{m}+\mathrm{v}=\frac{165}{2}$
Product $=\mathrm{mv}=1350$

$$\begin{gathered} \therefore \text{ Mean and Vairance be the roots of }{x}^2-82.5x+1350=0 \\ \begin{array}{r}\Rightarrow x^2-22.5x-60x+1350=0\\ \Rightarrow x(x-22.5)-60(x-22.5)=0\end{array} \\ \Rightarrow (x-22.5)(x-60)=0 \end{gathered}$$
On solving, 
$\begin{array}{l}\mathrm{m}=\mathrm{np}=60\mathrm{\&}\mathrm{v}=\mathrm{npq}=\frac{45}{2}\\ \therefore \mathrm{q}=\frac{3}{8}\therefore \mathrm{P}=\frac{5}{8}\end{array}$
Hence $\boxed{\mathrm{n}=96}$