Let $A=\{x\in N:1\le x\le 17\}$ and $B=\{ax+b:x\in X,a,b\in R,a>0\}.$

If mean and variance of elements 

of $B$ are 17 and 216 respectively, than $a+ b$ is equal to

We know that the mean and variance of first $n$

natural numbers are $\frac{n+1}{2}$ and $\frac{n^2-1}{12}$ respectively.

$\therefore \overline{X}=\frac{17+1}{2}=9$ and Var $\left(X\right)=\frac{{17}^2-1}{12}=24$

Now, $y_i=a{x}_i+b,i=1,2,\dots ,17$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\overline{Y}=a\overline{X}+b\text{ and }\mathrm{Var}\left(Y\right)=a^2\mathrm{Var}\left(X\right)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}17=9a+b\text{ and }216=a^2\left(24\right)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a=3,b=-10 [\because \mid z>0]\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a+b=-7\end{array}$