#### 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

`⇒ S_1 = \frac { n }{ 2 } {2a + (n – 1)d} ….(i)`

S_2 = Sum of 2n terms

`⇒ S_2 = \frac { 2n }{ 2 } [2a + (2n – 1) d] ….(ii)`

and, S_{3} = Sum of 3n terms

`⇒ S_3 = \frac { 3n }{ 2 } [2a + (3n – 1) d] ….(iii)`

Now, S_{2} – S_{1}

`= \frac { 2n }{ 2 } [2a + (2n – 1) d] – \frac { n }{ 2 } [2a + (n –1) d]`

`S_2 – S_1 = \frac { n }{ 2 } [2 {2a + (2n – 1)d} – {2a + (n – 1)d}]`

`= \frac { n }{ 2 } [2a + (3n – 1) d]`

`\∴ 3(S_2 – S_1 ) = \frac { 3n }{ 2 } [2a + (3n – 1) d] = S_3 ` [Using (iii)]

Hence, S_{3} = 3 (S_{2} – S_{1} )

Is there an error in this question or solution?

Solution The sum of n, 2n, 3n terms of an A.P. are S1 , S2 , S3 respectively. Prove that S3 = 3(S2 – S1 ) Concept: Sum of First n Terms of an AP.