Question

Let $X=\{x\in N:1\le x\le 17\}$ and $Y=\{ax+b:x\in X\text{and}a,b\in R,a>0\}$. If mean and variance of elements of $Y$ are $17$ and $216$ respectively then $a+b$ is equal to:

• A. $-27$
• B. $7$
• C. $-7$
• D. $9$

Answer 1

The correct option is C $-7$

$X:\{1,2,.........17\}$

$Y:\{ax+b:x\in X\&a,b\in R,a>0\}$ Given Var $\left(Y\right)=216$

$\frac{\sum y_1^2}{n}-{\left(\text{mean}\right)}^2=216$

$\Rightarrow \frac{\sum y_1^2}{17}-289=216$
$\Rightarrow \sum y_1^2=8585$
$\Rightarrow {(a+b)}^2+{(2a+b)}^2+....+{(17a+b)}^2=8585$
$\Rightarrow 105a^2+b^2+18ab=505\cdots \left(1\right)$

Now $\sum y_1=17\times 17$

$a(17\times 9)+17.b=17\times 17$

$9a+b=17\cdots \left(2\right)$
from equation $\left(1\right)\&\left(2\right)$

$a=3,b=-10$
$a+b=-7$