Question

# The sum of $$n, 2n, 3n$$ terms of an A.P. are $$S_1, S_2, S_3$$ respectively. Prove that $$S_3 = 3 (S_2 - S_1)$$

Solution

## Let $$a$$ be the first term and $$d$$ be the common difference of the given A.P. Then,$$S_1 =$$ Sum of $$n$$ terms $$\Longrightarrow S_1 = \dfrac{n}{2} \{2a + (n - 1)d \}$$                ...(i)$$S_2 =$$ Sum of $$2n$$ terms $$\Longrightarrow S_2 = \dfrac{2n}{2} \{2a + (2n - 1)d \}$$          ...(ii) and, $$S_3 =$$ Sum of $$3n$$ terms $$\Longrightarrow S_3 = \dfrac{3n}{2} \{2a + (3n - 1)d \}$$        ...(iii)Now, $$S_2 - S_1 = \dfrac{2n}{2} \{2a + (2n - 1)d \} - \dfrac{n}{2} \{2a + (n - 1)d \}$$$$\Longrightarrow S_2 - S_1 = \dfrac{n}{2} [2 \{2a + (2n - 1)d \} - \{2a + (n - 1)d \}] = \dfrac{n}{2} \{2a + (3n - 1)d \}$$$$\therefore 3(S_2 - S_1) = \dfrac{3n}{2} \{2a + (3n - 1)d \} = S_3$$            [Using (iii)]Hence,  $$S_3 = 3 (S_2 - S_1)$$Mathematics

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