The sum and product of the mean and variance of a binomial distribution are 82.5 and 1350 respectively then the number of trailas in the binomial distribution is

Mean $\mu =np$

$veriance{\sigma }^2=npq$

$np+npq=82.5\to \left(1\right)np.npq=1350$

${\left(np-npq\right)}^2={\left(np+npq\right)}^2-4\left(np\right).\left(npq\right)$

$={\left(82.5\right)}^2-4\left(1350\right)$

$=6806.25-5400$

$=1406.25={\left(37.5\right)}^2$

$np-npq=37.5\to \left(2\right)$

$\left(1\right)+\left(2\right)\Rightarrow 2np=120\Rightarrow np=60$

$From\left(2\right)npq=np-37.5=60-37.5=22.5$

$\therefore np=60,npq=\frac{45}{2}$

$q=\frac{npq}{np}=\frac{45}{2\times 60}=\frac{3}{8},p=1-q=1-\frac{3}{8}=\frac{5}{8}$

$np=60\Rightarrow n\left(\frac{5}{8}\right)=60$

$\Rightarrow n=96$