If $S_{r}$ denotes the sum of the first $r$ terms of an A.P, then $S_{3 n} : (S_{2 n} - S_{n})$ is

n

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3 n

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3

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None of these

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Solution

The correct option is D $3$
We know $s_{n} = \frac{n}{2} [2 a + n - 1 \times d]$
$\frac{s_{n}}{s_{2 n} - s_{n}} = \frac{\frac{3 n}{2} [2 a + 3 n - 1 \times d]}{\frac{2 n}{2} [2 a + 3 n - 1 \times d] - \frac{n}{2} [2 a + n - 1 \times d]}$
$= \frac{\frac{3 n}{2} [2 a + 3 n - 1 \times d]}{\frac{n}{2} [(4 a + 2 d) (2 n - 1)] [2 a + n - 1 \times d]}$
$= \frac{\frac{3 n}{2} [2 a + 3 n - 1 \times d]}{\frac{n}{2} [(4 a + 2 d) (2 n - 1) - 2 a - n - 1 \times d]}$
$= \frac{3 [2 a + 3 n - 1 d]}{(2 a + 2 d) (2 n - 1) - n - 1 d}$
$= \frac{(6 a + 3 d) (3 n - 1)}{(2 a + d) 4 n - 2 - n + 1}$
$= \frac{6 a + 9 n d - 3 d}{2 a + 3 n d - d}$
$= \frac{3 (2 a + 3 n d - n)}{2 a + 3 n d - d}$
$= 3$

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Similar questions

If S_rdenotes the sum of the first r terms of an AP . Then find

S_3n: (S_2n- S_n)

S_{r}

S_{3 n} : (S_{2 n} - S_{n})

n

1, 2, 3, 4, . . . . . .

S_{2 n} = 3 S n

\frac{S_{3 n}}{S n} = ?

S_{r}

\frac{S_{3 r}}{S_{2 r} - S_{r}}

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