Let the sum of the first n terms of a non-constant A . P . ,   a 1 ,   a 2 ,   a 3 ,   .…

Let the sum of the first n terms of a non-constant A.P., a1, a2, a3, ...., an be 50n+n(n-7)2A, where A is a constant. If d is the common difference of this A.P., then the ordered pair d, a50 is equal to
  1. 50, 50+46A
  2. A, 50+45A
  3. 50, 50+45A
  4. A, 50+46A

Solution

Given sum of n terms is Sn=50n+nn-72A   (1)

Sn-1=50(n-1)+(n-1)n-82A   2

Subtracting (1) and (2), we get

Sn-Sn-1=50n-50n-1+nn-72A-n-1n-82A

Sn-Sn-1=50n-n+1+A2nn-7-n-1n-8

Sn-Sn-1=50+A2n2-7n-n2+9n-8

Tn=Sn-Sn-1=50+A(n-4)

Hence, T1=50-3A and T2=50-2A

d=T2-T1=A and T50=50+46A.

Asked in: JEE Main 2019 (09 Apr Shift 1)

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