Let x_1,x_2,… ,x_100 be in an arithmetic progression, with x_1=2 and their mean equal to 200 . If y_i=i(x_i-i),1≤ i≤ 100, then the mean of y_1,y_2,…,…

Given,

Mean of ${x}_{1},{x}_{2}..........,{x}_{100}$x1​,x2​..........,x100​ is $200$200

So, the sum of observation will be,

$$\sum {x}_{i}=100\times 200$$∑xi​=100×200

Now using the sum of $A.P$A.P formula in above equation as all terms are in arithmetic progression we get,

$$\frac{100}{2}({x}_{1}+{x}_{100})=100\times 200$$2100​(x1​+x100​)=100×200

$$\Rightarrow 50(2+{x}_{100})=100\times 200$$⇒50(2+x100​)=100×200

$$\Rightarrow {x}_{100}=398$$⇒x100​=398

$$\Rightarrow {x}_{1}+99d=398$$⇒x1​+99d=398

$$\Rightarrow d=4$$⇒d=4

Now, ${x}_{i}=2+(i-1)4=4i-2$xi​=2+(i−1)4=4i−2

So, ${y}_{i}=i({x}_{i}-i)=3{i}^{2}-2i$yi​=i(xi​−i)=3i2−2i

Now finding mean we get,

$$\bar{y}=\frac{1}{100}\sum {y}_{i}$$yˉ​=1001​∑yi​

$$\Rightarrow \bar{y}=\frac{1}{100}\sum (3{i}^{2}-2i)$$⇒yˉ​=1001​∑(3i2−2i)

$$\Rightarrow \bar{y}=\frac{1}{100}[3\times \frac{100\times 101\times 201}{6}-2\frac{100\times 101}{2}]$$⇒yˉ​=1001​[3×6100×101×201​−22100×101​]

$$\Rightarrow \bar{y}=[\frac{101\times 201}{2}-101]$$⇒yˉ​=[2101×201​−101]

$$\Rightarrow \bar{y}=10049.5$$⇒yˉ​=10049.5