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Answer 1

Answer:

Given :

• S₁ , S₂ , S₃ are the sum of n terms of 3 arithmetic series.

• first term of each series is 1

• common difference of the respective series are 1 , 2 , 3

To prove :

S₁ + S₃ = 2S₂

Formula :

Sum of n terms in an AP., is given by,

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

Solution :

Let's find the sum of n terms of each arithmetic series.

Sum of n terms of first series :

first term, a = 1

common difference, d = 1

 $\sf S_1=\dfrac{n}{2}[2(1)+(n-1)(1)] \\\\ \sf S_1=\dfrac{n}{2}[2+n-1] \\\\ \sf S_1=\dfrac{n}{2}[n+1]$

Sum of n terms of second series :

first term, a = 1

common difference, d = 2

  $\sf S_2=\dfrac{n}{2}[2(1)+(n-1)(2)] \\\\ \sf S_2=\dfrac{n}{2}[2+2n-2] \\\\ \sf S_2=\dfrac{n}{2}[2n] \\\\ \sf S_2=n^2$

Sum of n terms of third series :

first term, a = 1

common difference, d = 3

 $\sf S_3=\dfrac{n}{2}[2(1)+(n-1)(3)] \\\\ \sf S_3=\dfrac{n}{2}[2+3n-3] \\\\ \sf S_3=\dfrac{n}{2}[3n-1]$

we have to prove S₁ + S₃ = 2S₂

LHS = S₁ + S₃

$\sf S_1+S_3=\dfrac{n}{2}[n+1]+\dfrac{n}{2}[3n-1] \\\\ \sf S_1+S_3=\dfrac{n}{2} [n+1+3n-1] \\\\ \sf S_1+S_3=\dfrac{n}{2} [4n] \\\\ \sf S_1+S_3=2n^2$

RHS = 2S₂

$\sf 2S_2=2 \times n^2 \\\\ \sf 2S_2=2n^2$

LHS = RHS

S₁ + S₃ = 2S₂

Hence proved!