1. Interpret the sets

Given $X = \{x \in \mathbb{N} : 1 \le x \le 17\} = \{1,2,3,\dots,17\}.$

Also, $Y = \{ax+b : x \in X,\ a,b \in \mathbb{R},\ a>0\}.$

So the elements of $Y$ are obtained by applying the linear transformation $y = ax+b$ to each element of $X$.

2. Mean and variance of $X$

Since $X = \{1,2,3,\dots,17\}$ is an arithmetic progression:

Mean of $X$

$\bar X = \frac{1+17}{2} = 9.$

Variance of $X$

For the numbers $1,2,\dots,n$, $\operatorname{Var}(X)=\frac{n^2-1}{12}.$ Here $n=17$, so \operatorname{Var}(X)=\frac{17^2-1}{12}=rac{289-1}{12}=rac{288}{12}=24.

Thus, $\mu_X=9, \qquad \sigma_X^2=24.$

3. Use transformation formulas

If $Y = aX+b,$ then

• mean transforms as $\mu_Y = a\mu_X + b,$
• variance transforms as $\sigma_Y^2 = a^2\sigma_X^2.$

Given: $\mu_Y = 17, \qquad \sigma_Y^2 = 216.$

So, $a^2 \cdot 24 = 216.$ Hence, $a^2 = 9.$ Since $a>0$, $a=3.$

Now use the mean: $a\mu_X + b = 17$ $3\cdot 9 + b = 17$ $27+b=17$ $b=-10.$

4. Find $a+b$

$a+b = 3 + (-10) = -7.$

5. Check options

• A: $7$ ❌
• B: $9$ ❌
• C: $-7$ ✅
• D: $-27$ ❌

Therefore, the correct answer is $\boxed{-7}.$