Let the sum of n, 2n, 3n terms of an A.P be S₁, S₂, S₃ respectively. Show that S₃ = 3(S₂ - S₁)

Solution:

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

Therefore,

S₁ = n/2 [2a + (n - 1) d] ....(1)

S₂ = 2n/2 [2a + (2n - 1) d]

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

S₃ = 3n/2 [2a + (3n - 1) d] ....(3)

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

S₂ - S₁ = 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(S₂ - S₁) = 3n/2 [2a + (3n - 1) d)]

Hence, S₃ = 3(S₂ - S₁) proved

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NCERT Solutions Class 11 Maths Chapter 9 Exercise ME Question 3

Let the sum of n, 2n, 3n terms of an A.P be S₁, S₂, S₃ respectively. Show that S₃ = 3(S₂ - S₁)

Summary:

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