Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that S3 = 3 (S2– S1)
Let a and b be the first term and the common difference of the A.P. respectively.
Therefore,
From (1) and (2), we obtain
Hence, the given result is proved.
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3 Comment(s) on this Question
Awesome sir
What a nice answer it is!
how about if we solve this question like this... S1 = n S2 = n + 2n = 3n S3 = n + 2n + 3n = 6n according to question, S3 = 3(S2 -S1 ) by putting the value of S1, S2, S3 6n = 3(3n - n) 6n = 3(2n) 6n = 6n so, LHS = RHS Hence Proved
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