Question

The sum of the first n terms of an AP is 2n. The sum of the first 2n terms is 3n.find the sum of the first 3n terms.

Answer 1

Let the first term of AP be 'a' and common difference be 'd'

Thus sum of n terms = (n/2)(2a+(n-1)d) = 2n

2a + (n-1)d =4 -----------------(1)

Now Sum of 2n terms = (2n/2)(2a + (2n-1)d) =3n

2a + (2n-1)d = 3 ----------------(2)

Subtracting eq(1) from eq(2) we get

nd = -1 -----------------(3)

Now Sum of first 3n terms = (3n/2)(2a+(3n-1)d) = (3n/2)[(2a+(2n-1)d)+nd] = (3n/2)[3 +nd] = (3n/2)(3-1)=3n

[Since (2a + (2n-1)d ) = 3 fro eq (2) and nd =-1 from eq (3)]

Thus Sum of first 3n terms = 3n