Question

The sums of first n terms of three arithmetic progressions are S1, S2 and S3 respectively. The first term of each A.P. is 1 and their common differences are 1,2 and 3 respectively. Prove that S2 + S3 = 2Sr
Please give me the answer as soon as possible ​

Answer 1

Answer:-

Given:

Sum of first n terms of three AP's are S₁ , S₂ , S₃.

First terms of three sequences = 1

Common difference of three sequences = 1 , 2 , 3.

We know that,

Sum of first n terms of an AP (Sₙ) = n/2 * [ 2a + (n - 1)d ]

So,

For First AP :

★ S₁ = n/2 * [ 2(1) + (n - 1)(1) ]

⟶ S₁ = n/2 ( 2 + n - 1)

⟶ S₁ = n(n + 1)/2 -- equation (1)

For Second AP :

★ S₂ = n/2 * [ 2(1) + (n - 1)(2) ]

⟶ S₂ = n/2 * ( 2 + 2n - 2)

⟶ S₂ = n² -- equation (2)

For third AP :

★ S₃ = n/2 * [ 2(1) + (n - 1)(3) ]

⟶ S₃ = n/2 * [ 2 + 3n - 3 ]

⟶ S₃ = n(3n - 1) / 2 -- equation (3).

Now,

We have to prove that:

S₁ + S₃ = 2S₂

Putting the values from equations (1) , (2) and (3) we get,

⟶ n(n + 1) / 2 + n(3n - 1)/2 = 2 × n²

⟶ n/2 * [ n + 1 + 3n - 1 ] = 2n²

⟶ 4n = 2n² * (2/n)

⟶ 4n = 4n

Hence, Proved.

Answer 2

Answer:

♠️Question♠️

The sums of first n terms of three arithmetic progressions are S1, S2 and S3 respectively. The first term of each A.P. is 1 and their common differences are 1,2 and 3 respectively. Prove that S2 + S3 = 2Sr

♠️ AnswEr ♠️

We know that,

Sum of first n terms of A.P (S¹) = N/2 × [2a + (n - 1) d]

♠️ First a.p

S1 = N/2 × [2(1)+ (n - 1) (1)]

S1 = n /2 ( 2 + n -1)

S1 = n (n +1 ) / 2

♠️ Second A. P

S2 = N/2 × [2(1)+ (n - 1) (2)]

S2 = N/2 × (2 +2n -2)

S2 = N²

♠️ Third A. P

S3 = n/2 × [2(1) + (n-1)(3)]

S3 = n/2 × [ 2 + 3n -3]

S3 = n(3n -1) /2

♠️ To prove

S1 + S3 = 2S²

♠️ Putting values

n(n +1)/2 + n (3n -1) /2 × n²

n/2 × [ n+1+ 3n - 1] = 2n²

4n = 2n² × (2/n)

4n = 4n

♠️ Hence proved