If the sum of the first terms of the AP series is equal to the sum of the first terms of the AP series , then equals
- A.10
- B.12
- C.11
- D.13
Show answer and solution
Correct option: C
Solution
Sum of first 2n terms of 2,5,8,... (a=2, d=3):
S = (2n/2)[2·2 + (2n-1)·3] = n[4 + 6n - 3] = n(6n+1)
Sum of first n terms of 57,59,61,... (a=57, d=2):
S = (n/2)[2·57 + (n-1)·2] = (n/2)[114 + 2n - 2] = n(56+n)
Setting equal: n(6n+1) = n(56+n), and since n≠0, divide by n:
6n+1 = 56+n → 5n = 55 → n = 11
S = (2n/2)[2·2 + (2n-1)·3] = n[4 + 6n - 3] = n(6n+1)
Sum of first n terms of 57,59,61,... (a=57, d=2):
S = (n/2)[2·57 + (n-1)·2] = (n/2)[114 + 2n - 2] = n(56+n)
Setting equal: n(6n+1) = n(56+n), and since n≠0, divide by n:
6n+1 = 56+n → 5n = 55 → n = 11
[2001]