Progression (Sequence and Series)HardBloom L3
Question
Let $S_{n},S_{2n},S_{3n}$ are respectively the sums of first $n,2n,3n$ terms of an arithmetic progression, then $S_{3n} =$
Options
A.$2\left( {\text{ }S}_{2n} - S_{n} \right)$
B.$\frac{3}{2}\left( {\text{ }S}_{2n} - S_{n} \right)$
C.$3\left( {\text{ }S}_{2n} - S_{n} \right)$
D.$6\left( {\text{ }S}_{2n} - S_{n} \right)$
Solution
Sol. [Key Concept: With $S_{n},S_{2n},S_{3n}$ of an A.P., $S_{3n} = 3(S_{2n} - S_{n})$.]
$$S_{2n} - S_{n} = \frac{2n}{2}(2a + (2n - 1)d) - \frac{n}{2}(2a + (n - 1)d) = \frac{n}{2}(2a + (3n - 1)d)$$
$$= \frac{1}{3}{\text{ }S}_{3n}$$
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