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  • Let the sum of n, 2 n, 3 n terms of an A.P. be S 1, S 2 and S 3, respectively, show that S 3 = 3 (S 2 minus S 1)

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)

Answers (1)

best_answer

Let a be first term and d be common difference of AP.

S_n=\frac{n}{2}[2a+(n-1)d]=S_1..................................1

S_2n=\frac{2n}{2}[2a+(2n-1)d]=S_2..................................2

S_2_n=\frac{3n}{2}[2a+(3n-1)d]=S_3..................................3

Subtract equation 1 from 2,

S_2-S_1=\frac{2n}{2}[2a+(2n-1)d]-\frac{n}{2}[2a+(n-1)d]

                =\frac{n}{2}[4a+4nd-2d-2a-nd+d]

               =\frac{n}{2}[2a+3nd-d]

               =\frac{n}{2}[2a+(3n-1)d]

\therefore 3(S_2-S_1)=\frac{3n}{2}[2a+(3n-1)d]=S_3

Hence, the result is proved.

Posted by

seema garhwal

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