# Let the sum of n, 2n, 3n terms of an A.P be S_{1}, S_{2}, S_{3} respectively. Show that S_{3} = 3(S_{2} - S_{1})

**Solution:**

Let a and d be the first term and common difference of the A.P respectively.

Therefore,

S_{1} = n/2 [2a + (n - 1) d] ....(1)

S_{2} = 2n/2 [2a + (2n - 1) d]

= n [2a + (2n - 1) d] ....(2)

S_{3} = 3n/2 [2a + (3n - 1) d] ....(3)

By subtracting (1) and (2) , we obtain

S_{2} - S_{1} = n [2a + (2n - 1) d] - n/2 [2a + (n - 1) d]

= n [(4a + 4nd - 2d - 2a - nd + d)/2]

= n [(2a + 3nd - d)/2]

3(S_{2} - S_{1}) = 3n/2 [2a + (3n - 1) d)]

Hence, S_{3} = 3(S_{2} - S_{1}) proved

NCERT Solutions Class 11 Maths Chapter 9 Exercise ME Question 3

## Let the sum of n, 2n, 3n terms of an A.P be S_{1}, S_{2}, S_{3} respectively. Show that S_{3} = 3(S_{2} - S_{1})

**Summary:**

Thus, knowing that the sum of n, 2n, and 3n terms of the A.P be S1, S2, S3 we found out that S_{3} = 3(S_{2} - S_{1})