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Question

The sum of n terms of three arithmetical progression are S1, S2 and S3. The first term of each is unity and the common differences are 1, 2 and 3 respectively. Prove that S1 + S3 = 2S2.


Solution

We have,
S1 = Sum of n terms of an AP. with first term 1 and common difference 1

=(n2) [2 x 1+(n- 1) 1] =(n2) [n+ 1]
S2 = Sum of n terms of an AP. with first term 1 and common difference 2

=(n2)[2x1+(n—1)x2]= n2
S3 = Sum of n terms of an AP. with first term 1 and common difference3
=(n2) [2 x1+(n-1) × 3]=(n2)(3n-1)

Now, S1 + S3 = (n2) (n + 1) + (n2) (3n — 1)
= 2n2 and S2 =  n2

Hence S1 + S3 = 2S2


Mathematics

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