Question

If sums of n 2n and 3n terms of an AP are S1,S2 andS3 respectively then S3 /(S2 - S1)

0

1

2

3

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Answer 1

Given :

• Sum of first n terms of AP = S₁
• Sum of first 2n terms of AP = S₂
• Sum of first 3n terms of AP = S₃

To find :

$\sf \dfrac{S_3}{(S_2-S_1)}$

Formula used :

$\sf \: Sum \: of \: n \: term \: of \: AP \: = \dfrac{n}{2} \bigg( \: 2a + ( \: n \: - \: 1 \: )d \: \bigg)$

Here,

• n = number of terms
• a = 1st tern
• d = common terms

Solution :

$\sf \bullet \: S_1 = \dfrac{n}{2} \bigg( \: 2a + ( \: n \: - \: 1 \: )d \: \bigg) \\ \sf \bullet \: S_2 = \dfrac{2n}{2} \bigg( \: 2a + ( \:2 n \: - \: 1 \: )d \: \bigg) \\ \sf \bullet \: S_3 = \dfrac{3n}{2} \bigg( \: 2a + ( \:3 n \: - \: 1 \: )d \: \bigg)$

This gives,

$\sf \implies \: \dfrac{S_3}{(S_2-S_1)} \: = \: \dfrac{ \dfrac{3n}{2} \bigg( \: 2a + ( \:3 n \: - \: 1 \: )d \: \bigg)}{\dfrac{2n}{2} \bigg( \: 2a + ( \:2 n \: - \: 1 \: )d \: \bigg) \: - \: \dfrac{n}{2} \bigg( \: 2a + ( \: n \: - \: 1 \: )d \: \bigg) }$

Take (n/2) common in numerator and denominator

$\sf \implies \: \dfrac{S_3}{(S_2-S_1)} \: = \: \dfrac{ \dfrac{n}{2} \times 3 \times \bigg( \: 2a + ( \:3 n \: - \: 1 \: )d \: \bigg)}{ \bigg(\dfrac{n}{2} \bigg)\bigg( \: 2 \times (\: 2a + ( \:2 n \: - \: 1 \: )d \: ) \: \: - \:\: 2a + ( \: n \: - \: 1 \: )d \: \bigg) }$

$\sf \implies \: \dfrac{S_3}{(S_2-S_1)} \: = \: \dfrac{ 3\bigg( \: 2a + ( \:3 n \: - \: 1 \: )d \: \bigg)}{ \bigg( \: 4a +2 ( \:2 n \: - \: 1 \: )d \: \: \: - \:\: 2a + ( \: n \: - \: 1 \: )d \: \bigg) }$

$\sf \implies \: \dfrac{S_3}{(S_2-S_1)} \: = \: \dfrac{ 3\bigg( \: 2a + (3n \: - \: 1 \: )d \: \bigg)}{ \bigg( \: 4a + 4 n d\: - 2d \: - \: 2a - \: nd \: + d \: \bigg) }$

$\sf \implies \: \dfrac{S_3}{(S_2-S_1)} \: = \: \dfrac{3 \bigg( \:2 a + \:3nd \: - \: d \: \bigg)}{ \bigg( \: 2a + 3 n d\: - d \: \: \bigg) }$

cancel out common term from numerator and denominator

$\sf \implies \: \dfrac{S_3}{(S_2-S_1)} \: = \: 3$

ANSWER : 3