First slide

Arithmetic progression

Question

$Let S_{n} denote the sum of the first n terms of an A.P. if S_{2 n} = 3 S_{n} then S_{3 n} : S_{s} =$

Moderate

A

4
B

6
C

8
D

10

Solution

$\begin{array}{l} \frac{S_{2 n}}{S_{n}} = \frac{3}{1} so, \frac{\frac{2 n}{2} [2 a + (2 n - 1) d]}{\frac{n}{2} [2 a + (n - 1) d]} = \frac{3}{1} \\ \frac{4 a + (4 n - 2) d}{2 a + (n - 1) d} = 3 \\ 4 a + (4 n - 2) d = 6 a + (3 n - 3) d \\ (n + 1) d = 2 a \\ \frac{S_{3 n}}{S_{n}} = \frac{\frac{3 n}{2} [2 a + (3 n - 1) d]}{\frac{n}{2} [2 a + (n - 1) d]} = \frac{3 [(n + 1) + (3 n - 1)]}{[(n + 1) + (n - 1)]} = 3 \frac{(4 n)}{2 n} = 6 \end{array}$

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