If S 1- S 2- S 3 are the sum of n - 2 n - 3 n terms respectively of an A-P- then S 3- S 2- S 1A- 1B- 2C- 3D- 4

The correct option is C 3S_1 = n/2 [2a + (n 1)d], S_2 = n[2a + (2n 1)d] S_2 S_1 = na + (3n 1) nd/2 = n/2 [2a + (3n 1)d] S_3 = 3n/2 [2a + (3n 1)d ...

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Arithmetic Progression

If S 1, S 2, ...

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If $S_{1}, S_{2}, S_{3}$ are the sum of n, 2n, 3n terms respectively of an A.P. then $\frac{S_{3}}{S_{2} - S_{1}}$

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If sum of n, 2n, 3n terms of an A.P. are $S_{1}, S_{2}, S_{3}$ respectively, then the value of $\frac{S_{3}}{S_{2} - S_{1}}$ is

S_{1}, S_{2}

S_{3}

\frac{s_{3}}{(s_{2} - s_{1})}

, 2 n, 3 n

S_{1}, S_{2}, S_{3}

S_{3} = 3 (S_{2} - S_{1})

The sum of n, 2n, 3n terms of an A.P. are $S_{1}$ , $S_{2}$ , $S_{3}$ respectively. Prove that $S_{3}$ = 3( $S_{2}$ – $S_{1}$ ).

Let sum of n, 2n, 3n terms of an A.P. be $S_{1}, S_{2}$ and $S_{3}$ respectively, show that $S_{3} = 3 (S_{2} - S_{1})$

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