Question

The sum of n terms three AP are S1,S2,S3 respectively the first term of each AP is 1,2 and 3 respectively. Prove that S1+S2=2S3

Answer 1

S₁ + S₃  = 2S₂ if The first term of three AP's are 1 and the common difference are 1, 2 and 3 respectively.

Step-by-step explanation:

Correct Question is

The first term of three AP's are 1 and the common difference are 1, 2 and 3 respectively.

a = 1   d = 1

last term = 1 + (n - 1)1  = n

S₁ = (n/2)(1 + n) = n(n + 1)/2

a = 1   d = 2

last term = 1 + (n - 1)2  = 2n-1

S₂ = (n/2)(1 + 2n-1) = n²

a = 1   d = 3

last term = 1 + (n - 1)3  = 3n-2

S₃ = (n/2)(1 + 3n-2) = n(3n - 1)/2

S₁ + S₃  =  n(n + 1)/2  + n(3n - 1)/2

=> S₁ + S₃  =  (n/2)(n + 1 + 3n - 1)

=> S₁ + S₃  = (n/2) 4n

=> S₁ + S₃  = 2n²

=> S₁ + S₃  = 2S₂

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Answer 2

$\longrightarrow$$\large\tt\red{S=\frac{n}{2}(2a+(n-1)d)}$

$\longrightarrow$$\large\tt\purple{S1=\frac{n}{2}(2×1+(n-1)×1)}$

$\longrightarrow$$\large\tt\purple{\frac{n}{2}(2+(n-1)}$

$\longrightarrow$$\large\tt\purple{\frac{n(n+1)}{2}}$.....(1)

$\longrightarrow$$\large\tt\pink{S2=\frac{n}{2} (2 \times 1 + (n - 1)2)}$

$\longrightarrow$$\large\tt\pink{S2 = \frac{n}{2} (2 + 2n - 2)}$

$\longrightarrow$$\large\tt\pink{ \frac{n}{2} \times 2n = {n}^{2} }$.....(2)

$\longrightarrow$$\large\tt\green{S3=\frac{n}{2}(2×1+(n-1)3)}$

$\longrightarrow$$\large\tt\green{\frac{n}{2}(2+3n-3)}$

$\longrightarrow$$\large\tt\green{\frac{n}{2}(3n-1)}$.....(3)

$\large\tt\blue{LHS=S1+S3}$

$\longrightarrow$$\large\tt\orange{ \frac{n(n + 1)}{2} + \frac{n}{2} (3n - 1)}$

$\longrightarrow$$\large\tt\orange{ \frac{n}{2} (n + 1 + 3n - 1)}$

$\longrightarrow$$\large\tt\orange{\frac{n}{2} \times 4n = 2 {(n)}^{2} }$

$\therefore$$\longrightarrow$$\large\tt\orange{S1+S3=2S2}$