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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 81 of 106
Marks: +1, -0
Let the sum of n, 2n, 3n terms of an A.P. be S1,S2S_1, S_2 and S3S_3, respectively, show that S3S_3 = 3(S2−S1).3(S_2 - S_1).
Solution:  
Let the first term be a and the common difference be d.
According to question
S1S_1 = n2\frac{n}{2} [2a + (n - 1) d] , S2S_2 = 2n2\frac{2n}{2} [2a + (2n - 1) d] , S3S_3 = 3n2\frac{3n}{2} [2a + (3n - 1) d]
Now , S2−S1S_2 - S_1 = 2n2\frac{2n}{2} [2a + (2n - 1) d] - n2\frac{n}{2} [2a + (n - 1) d]
= n2\frac{n}{2} [(4a + (4n - 2) d) - (2a + (n - 1) d)] = n2\frac{n}{2} [(4a - 2a) + (4n - 2 - n + 1) d]
= n2\frac{n}{2} [2a + (3n - 1) d] = 13[3n2[2a+(3n−1)d]]\frac{1}{3}\left[\frac{3n}{2}[2a+(3n-1)d]\right]
⇒ S2−S1S_2 - S_1 = 13S3\frac{1}{3} S_3
Hence S3S_3 = 3(S2−S1)3(S_2 - S_1)
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