Let 1 x 1 ,   1 x 2 , … , 1 x n ( x i ≠ 0 for i = 1 ,   2 , … . ,   n ) be…

Let 1x1, 1x2,,1xn(xi0 for i=1, 2,., n) be in A.P. such that x1=4 and x21=20 . If n is the least positive integer for which xn>50, then i=1n1xi is equal to
  1. 3
  2. 18
  3. 134
  4.  138

Solution

Given 1x1,1x2,,1xn are in A.P. and x1=4 and x21=20.

a=14

Given, x21=20.

14+20·d=120

d=1100

Given, xn>50.

1xn<150

14- n-1100< 150n>24

 n=25

Now, i=1251xi=2522×14-1100×24=134

Asked in: JEE Main 2018 (16 Apr Online)

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