Let [x_1, x_2, , x_100] be in an arithmetic progression, with x_1=2 and their mean equal t…

Question

Let $$x_{1}, x_{2}, \ldots, x_{100}$$ be in an arithmetic progression, with $x_{1}=2$ and their mean equal to 200 . If $$y_{i}=i\left(x_{i}-i\right), 1 \leq i \leq 100$$, then the mean of $$y_{1}, y_{2}, \ldots, y_{100}$$ is :

Accepted Answer

We have, mean of ${x}_{1},{x}_{2}\ldots \ldots {x}_{100}=200$

Where, ${x}_{1},{x}_{2}\ldots {x}_{100}$ are in AP with first term as 2.

$\begin{gathered} =200 \ =\frac{\sum _{i=1}^{100}{x}_{i}}{100}=200 \end{gathered}$

$\begin{gathered} \frac{100}{2}	imes [2	imes 2+99d]=20000 \ \Rightarrow 4+99d=400 \ \Rightarrow 99d=396 \ d=4 \end{gathered}$

Also,

$\begin{gathered} {y}_{i} =i({x}_{i}-i) \ \ =i[2+(i-1)4-i] \ \ =i[3i-2] \ \ =3{i}^{2}-2i \end{gathered}$

$\begin{gathered} =\frac{\sum _{i=1}^{100}{y}_{i}}{100} \ =\frac{1}{100}[\sum _{i=1}^{100}(3{i}^{2}-2i)] \ =\frac{1}{100}[\frac{3	imes 100	imes 101	imes 201}{6}-2	imes \frac{100	imes 101}{2}] \ =\frac{20301}{2}-101 \ =10049.50 \end{gathered}$

Solutions · 1