QJEEM21Maad88: JEE Main 2021 Mathematics - Class 11 JEEM21Maad88 Solution: Let S 1 be the sum of first 2n terms of an arithmetic…

Question

Verified Solution • 2025–26 Session moderate

Let S₁ be the sum of first 2n terms of an arithmetic progression. Let S₂ be the sum of first 4n terms of the same arithmetic progression. If (S₂

-

S₁) is 1000, then the sum of the first 6n terms of the arithmetic progression is equal to :

Accepted Answer

S₁ =

{{2n} \over 2}

[2a + (2n

-

1)d]

S₂ =

{{4n} \over 2}

[2a + (4n

-

1)d]

(where a = T₁ and d is common difference)

S₂

-

S₁

\Rightarrow

2n[2a + (4n

-

1)d]

-

n[2a + (2n

-

1)d] = 1000

\Rightarrow

n[2a + d(8n

-

2

-

2n + 1)] = 1000

\Rightarrow

n[2a + (6n

-

1)d] = 1000

S₆ =

{{6n} \over 2}

[2a + (6n

-

1)d] = 3(S₂

-

S₁) = 3000

Step-by-Step Verified Solution