Question

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

Answer 1

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

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:
\[S = \dfrac{n}{2}(2a + (n - 1)d)\]
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:
\[{S_1} = \dfrac{n}{2}(2(1) + (n - 1)(1))\]
Simplifying, we have:
\[{S_1} = \dfrac{n}{2}(2 + n - 1)\]
\[{S_1} = \dfrac{n}{2}(1 + n).............(1)\]
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:
\[{S_2} = \dfrac{n}{2}(2(1) + (n - 1)(2))\]
Simplifying, we have:
\[{S_2} = \dfrac{n}{2}(2 + 2n - 2)\]
\[{S_2} = \dfrac{n}{2}(2n)\]
\[{S_2} = {n^2}...........(2)\]
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:
\[{S_3} = \dfrac{n}{2}(2(1) + (n - 1)(3))\]
Simplifying, we have:
\[{S_3} = \dfrac{n}{2}(2 + 3n - 3)\]
\[{S_3} = \dfrac{n}{2}(3n - 1)...........(3)\]
Adding equation (1) and equation (3), we have:
\[{S_1} + {S_3} = \dfrac{n}{2}(1 + n) + \dfrac{n}{2}(3n - 1)\]
Taking \[\dfrac{n}{2}\] as a common term, we have:
\[{S_1} + {S_3} = \dfrac{n}{2}(1 + n + 3n - 1)\]
Simplifying, we have:
\[{S_1} + {S_3} = \dfrac{n}{2}(4n)\]
\[{S_1} + {S_3} = 2{n^2}\]
From equation (2), we have:
\[{S_1} + {S_3} = 2{S_2}\]
Hence, we proved

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