Question

The sums of the first nth terms of 3APs are S1,S2 and S3 respectively.The first term of its AP is 1 and their common difference are 1,2 and 3 respectively.Prove that S1+S3=2S2

Answer 1

let a be the first term and d the common difference
A.T.Q.
S1 = n/2 [2a + (n-1)d1]
=n/2 [2 + (n-1)] (putting a as 1 and d as 1)
= n/2 [n+1]

S2= n/2 [2a + (n-1)d2]
= n/2 [2 + (n-1)2] (putting a as 1 and d as 2)
= n[1+ (n-1)]
=n²

S3= n/2 [2a + (n-1)d3]
=n/2 [2+ (n-1)3] (putting a as 1 and d as 3)
= n/2[3n-1]

to prove _ S1+S3=2S2
S1+S2
= n/2 [n+1] + n/2[3n-1]
=n/2[n+1+3n-1]
=n/2[4n]
=2n²
=>2S2