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Question: The sum of n terms of three arithmetic progressions are \[{S_1}\], \[{S_2}\] and \[{S_3}\]. The firs...

The sum of n terms of three arithmetic progressions are S1{S_1}, S2{S_2} and S3{S_3}. The first term of each is unity and the common differences are 1, 2, and 3 respectively. Prove that S1+S3=2S2{S_1} + {S_3} = 2{S_2}.

Explanation

Solution

Hint: The sum of n terms of an AP is given by the formula S=n2(2a+(n−1)d)S = \dfrac{n}{2}(2a + (n - 1)d) where a is the first term and d is the common difference. Use this formula to find S1{S_1}, S2{S_2} and S3{S_3} and prove that they satisfy the given condition S1+S3=2S2{S_1} + {S_3} = 2{S_2}.

Complete Complete step by step answer:
An arithmetic progression (AP) is a sequence of numbers such that each consecutive term differs by a constant number. This constant number is called the common difference of the AP. The AP is completely described with its first term and the common difference.
The sum of n terms of an AP where a is the first term and d is the common difference is given by the formula as follows:
S=n2(2a+(n−1)d)S = \dfrac{n}{2}(2a + (n - 1)d)
It is given that the first AP has the first term as 1 and the common difference is 1. The sum of n terms of this AP is given as follows:
S1=n2(2(1)+(n−1)(1)){S_1} = \dfrac{n}{2}(2(1) + (n - 1)(1))
Simplifying, we have:
S1=n2(2+n−1){S_1} = \dfrac{n}{2}(2 + n - 1)
S1=n2(1+n).............(1){S_1} = \dfrac{n}{2}(1 + n).............(1)
It is given that the second AP has the first term as 1 and the common difference is 2. The sum of n terms of this AP is given as follows:
S2=n2(2(1)+(n−1)(2)){S_2} = \dfrac{n}{2}(2(1) + (n - 1)(2))
Simplifying, we have:
S2=n2(2+2n−2){S_2} = \dfrac{n}{2}(2 + 2n - 2)
S2=n2(2n){S_2} = \dfrac{n}{2}(2n)
S2=n2...........(2){S_2} = {n^2}...........(2)
It is given that the third AP has the first term as 1 and the common difference is 3. The sum of n terms of this AP is given as follows:
S3=n2(2(1)+(n−1)(3)){S_3} = \dfrac{n}{2}(2(1) + (n - 1)(3))
Simplifying, we have:
S3=n2(2+3n−3){S_3} = \dfrac{n}{2}(2 + 3n - 3)
S3=n2(3n−1)...........(3){S_3} = \dfrac{n}{2}(3n - 1)...........(3)
Adding equation (1) and equation (3), we have:
S1+S3=n2(1+n)+n2(3n−1){S_1} + {S_3} = \dfrac{n}{2}(1 + n) + \dfrac{n}{2}(3n - 1)
Taking n2\dfrac{n}{2} as a common term, we have:
S1+S3=n2(1+n+3n−1){S_1} + {S_3} = \dfrac{n}{2}(1 + n + 3n - 1)
Simplifying, we have:
S1+S3=n2(4n){S_1} + {S_3} = \dfrac{n}{2}(4n)
S1+S3=2n2{S_1} + {S_3} = 2{n^2}
From equation (2), we have:
S1+S3=2S2{S_1} + {S_3} = 2{S_2}
Hence, we proved

Note: We can also start the proof from 2S22{S_2} and can rewrite the term in a manner such that we can get the expression for S1{S_1} and S3{S_3}. But it is easier and straight forward to go from S1+S3{S_1} + {S_3} to 2S22{S_2}.