Let x₁, x₂, , x₁₅ be 15 numbers in an arithmetic progression. If _ i=1 ¹⁵ (x_i - 10) = 30 and _ i=1 ¹⁵ (x_i - 10)^2 = 2580 , then the value of x₁^2 + x₁₅^2 is equal to

Step-by-step solution

Let y_i = x_i - 10 . We are given: _ i=1 ¹⁵ y_i = 30 and _ i=1 ¹⁵ y_i^2 = 2580 The mean of y_i is y = 30 15 = 2 . Thus, the mean of x_i is x = y + 10 = 12 . The variance of x_i is the same as the variance of y_i : ^2 = y_i^2 15 - ( y )^2 ^2 = 2580 15 - (2)^2 = 172 - 4 = 168 For an arithmetic progression of n=15 terms with common difference d , the variance is: ^2 = n^2 - 1 12 d^2 15^2 - 1 12 d^2 = 168 224 12 d^2 = 168 56 3 d^2 = 168 d^2 = 168 3 56 = 9 In an A.P. of 15 terms, the mean x is the middle term, which is