Question

If sn denotes the sum of first n terms of an ap then show that the common difference d of ap is d=sn-2sn-1+sn-2

Answer 1

The common difference d is defined as the difference between any two consecutive terms of the AP.

The sum of an AP is given as,

Sₙ = T₁ + T₂..................+ Tₙ₋₂ + Tₙ₋₁ + Tₙ

Also

Sₙ₋₁ = T₁ + T₂..................+ Tₙ₋₂ + Tₙ₋₁

hence , Sₙ can be written as,

Sₙ = Sₙ₋₁ + Tₙ

=> Tₙ = Sₙ - Sₙ₋₁

Similarly, Tₙ₋₁ can be written as

Tₙ₋₁ = Sₙ₋₁ - Sₙ₋₂

Now we know that,

d = Tₙ - Tₙ₋₁

=> d = (Sₙ - Sₙ₋₁) - (Sₙ₋₁ - Sₙ₋₂)

=> d = Sₙ - 2Sₙ₋₁ + Sₙ₋₂

Which is the required equation to be proved.

Answer 2

Answer:

Given

Sn = (n/2)[ 2a + ( n -1) d]

Now

=>  Sn - 2Sn-1 + Sn +  2

=> (n/2)[ 2a + ( n -1) d] - 2(n - 1)/2)[ 2a + ( n - 1 -1) d] + (n +2) / 2)[ 2a + ( n + 2 -1) d]

=> (1/2)[ 2an + n( n -1) d] + [ 4a(n - 1) + 2(n - 1)( n - 2) d] +[ 2a(n + 2)+ ( n + 1) (n + 2) d]

=> (1/2)[ 2a[ n - 2n + 2 + n + 2] + d [ n2 - n - 2n2 + 6n - 4 + n2 + 3n + 2] ]

=> (1/2)[ 2a(4) + d(8n - 2) ]

=> [ 4a + (4n - 1)d]

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@GauravSaxena01