Let x 1 , x 2 , x 3 , … . . , x 20 be in geometric progression with x 1 = 3 and the common ration 1 2 . A new data is constructed replacing each x i by x i - i 2 . If x is the mean of new data, then the greatest integer less than or equal to x is

Step-by-step solution

Given, x 1 , x 2 , x 3 , … . . , x 20 are in G.P with common difference 1 2 and first term as 3 Now using the formula for sum of G.P we get, ∑ i = 1 20 x i = 3 1 - 1 2 20 1 - 1 2 = 6 1 - 1 2 20 Now new sum of G.P with new data will be = ∑ i = 1 20 x i - i 2 = ∑ i = 1 20 x i 2 + i 2 - 2 x i i Now ∑ i = 1 20 x i 2 = 9 1 - 1 4 20 1 - 1 4 = 12 1 - 1 2 40 And ∑ i = 1 20 i 2 = 1 6 × 20 × 21 × 41 = 2870 And ∑ i = 1 20 x i i = 3 + 2 . 3 1 2 + 3 . 3 1 2 2 + 4 . 3 1 2