Given:

Sum of 2n terms of AP = 3(Sum of n terms of AP) 

Concept:

Arithmetic Progression: An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the previous term, except for the first term. This fixed number is called the common difference of the AP, which can be positive, negative, or zero.

Consider the series where the first term is a and the common difference is d.

a, a + d, a + 2d, ...........a + (n - 1) d

The sum of the nth term of AP is given by,

\(S_n = \frac{n}{2}[2a + (n - 1)d]\)

Calculation:

Let us consider the series where the first term is a and the common difference is d.

a, a + d, a + 2d, ...........a + (n - 1) d

The sum of the nth term of AP is given by,

\(S_n = \frac{n}{2}[2a + (n - 1)d]\)               .......(1)

The sum of the 2n th term of AP is given by,

\(S_{2n} = \frac{2n}{2}[2a + (2n - 1)d]\)            .....(2)

The sum of the nth term of AP is given by,

\(S_{3n} = \frac{n}{2}[2a + (3n - 1)d]\)           .......(3)

According to the question,  S2n = 3Sn

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

⇒ 4a + 4nd - 2d = 6a + 3nd - 3d

⇒ 2a = nd +d                     ......(4)

From equations (1) and (3)

\(\frac{S_{3n}}{S_n} = \frac{\frac{3n}{2}[2a + (3n - 1)d]}{\frac{n}{2}[2a + (n - 1)d]}\)

\(⇒ \frac{S_{3n}}{S_n} = \frac{3[nd + d + 3nd - d]}{[nd + d + nd - d]}\)

\(⇒ \frac{S_{3n}}{S_n} =6\)

 Shortcut Trick

Let n = 1

We know that,

S₁ = Sum of 1 term = First term of AP = a

S₂ = a + a + d  = 2a + d

S₃ = a + a + d + a+ 2d = 3a + 3d

According to the question  S2n = 3Sn that is S₂ = 3S₁

⇒ 2a + d = 3a 

⇒ a = d

\(\frac{S_3}{S_1} = \frac{3a + 3d}{a}\)

⇒ S₃/S₁ = 6