Question

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

$-7$

Explanation for correct answer:

Step-1 Mean of Y:

Given, $X=\left\{x\in N:1\le x\le 17\right\}$,$\Rightarrow X=\left\{1,2,3,...,17\right\}$

$Y=\left\{ax+b:x\in X\text{and}a,b\in R,a>0\right\}$

Mean of elements in $Y=E\left(Y\right)=17$

$$\begin{gathered} \sum _{i-1}^{17}{y}_i=\underset{i-1}{\overset{17}{\frac{1}{17}\sum }}\left(a{x}_i+b\right) \\ =\frac{1}{17}\left(\underset{i-1}{\overset{17}{a\sum }}{x}_i+b\underset{i-1}{\overset{17}{\sum 1}}\right) \\ =\frac{1}{17}(\left(1+2+3+...+17)a+17b\right)\left[\because \underset{i=1}{\overset{n}{\sum 1}}=n\right] \\ =\frac{1}{17}\left(\left(\frac{17\times 18}{2}\right)a+17b\right)\left[\because 1+2+...+n=\frac{n(n+1)}{2}\right] \\ =\frac{17}{17}\left(9a+b\right) \end{gathered}$$

$\Rightarrow$ $9a+b=17\to \left(i\right)$

Step-2 : Varinace of Y:

A variance of the elements in

$Y=E\left(Y^2\right)-{\left(E\left(Y\right)\right)}^2=216$

$\begin{array}{rcl}\Rightarrow E\left(Y^2\right)-{\left(17\right)}^2& =& 216\\ \Rightarrow \frac{{\displaystyle \sum _{i=1}^{17}}{\left(y_i\right)}^2}{17}& =& 216+289\\ \Rightarrow \sum _{i=1}^{17}{\left(y_i\right)}^2& =& 505\times 17\\ \Rightarrow \sum _{i=1}^{17}{\left(a{x}_i+b\right)}^2& =& 8585\\ \Rightarrow \sum _{i=1}^{17}{\left(a{x}_i\right)}^2+2ab\sum _{i=1}^{17}{x}_i+\underset{i=1}{\overset{17}{\sum {\left(b\right)}^2}}& =& 8585\end{array}$

$\Rightarrow {\mathrm{a}}^2(1^2+2^2+...+{17}^2)+2\mathrm{ab}(1+2+...+17)+17{\mathrm{b}}^2=8585\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }\underset{\mathbf{i}\mathbf{=}\mathbf{1}}{\overset{\mathbf{n}}{\mathbf{}\mathbf{}\mathbf{\sum }\mathbf{1}}}\mathbf{=}\mathbf{n}\mathbf{]}$

Using these formula$\mathbf{}{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{2}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathbf{n}}^{\mathbf{2}}\mathbf{}\mathbf{=}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{6}}\mathbf{;}\mathbf{}\mathbf{}\mathbf{1}\mathbf{+}\mathbf{2}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}\mathbf{n}\mathbf{=}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{2}}$

we get,

$$\begin{gathered} a^2\left(\frac{17\times 18\times 35}{6}\right)+\cancel{2}ab\left(\frac{17\times 18}{\cancel{2}}\right)+17b^2=17\times 505 \\ \cancel{17}\left(105a^2+18ab+b^2\right)=505\times \cancel{17} \\ \left(105a^2+18ab+b^2\right)=505\to \left(ii\right) \end{gathered}$$

Solving equations $\left(i\right)$ and $\left(ii\right)$ we get

$$\begin{gathered} \left(81a^2+18ab+b^2\right)+24a^2=505 \\ {\left(9a+b\right)}^2+24a^2=505 \\ {\left(17\right)}^2+24a^2=505 \\ 24a^2=505-289 \\ {a}^2=\frac{216}{24} \\ {a}^2=9 \\ a=3 \end{gathered}$$

Substituting $a=3$ in equation $\left(i\right)$

$$\begin{gathered} 9\times 3+b=17 \\ 27+b=17 \\ b=17-27 \\ =-10 \end{gathered}$$

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

Hence, the correct answer is option (C).