Question

The sum of first n terms of three A.P s are ${S_1},{S_2},{S_3}$. The first term of each is 5 and their common differences are 2, 4, and 6 respectively. Prove that ${S_1} + {S_3} = 2{S_2}$.

Answer 1

Hint: In this type of question, we will find the common difference by using the sum of $n$ terms the arithmetic progression which is, \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\], where a is the first term, d is the common difference, apply this formula for each series.

Complete step-by-step answer:
Given that the sum of first n terms of three A.P s are ${S_1},{S_2},{S_3}$, and the first term of each is 5 and their common differences are 2, 4, and 6 respectively.
Let us consider that ${S_1},{S_2},{S_3}$ are the sums of $n$ three series in arithmetic progression.
We know that the arithmetic progression is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value.
Now we will take ${S_1}$, we get,
Now using the formula of the sum of n terms of the arithmetic progression, A.P i.e., ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$, here a=5, d=2,
Now substituting the values we get,
${S_1} = \dfrac{n}{2}\left[ {2\left( 5 \right) + \left( {n - 1} \right)2} \right]$,
Now simplifying we get,
${S_1} = \dfrac{n}{2}\left[ {10 + 2n - 2} \right]$
$ \Rightarrow {S_1} = \dfrac{n}{2}\left[ {2n + 8} \right]$,
$ \Rightarrow {S_1} = n\left[ {n + 4} \right]$,
$ \Rightarrow {S_1} = {n^2} + 4n$,
Now considering ${S_2}$ and again using the formula we get,
Now using the formula of the sum of n terms of the arithmetic progression, A.P i.e., ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$, here a=5, d=4,
Now substituting the values we get,
${S_2} = \dfrac{n}{2}\left[ {2\left( 5 \right) + \left( {n - 1} \right)4} \right]$,
Now simplifying we get,
${S_2} = \dfrac{n}{2}\left[ {10 + 4n - 4} \right]$
$ \Rightarrow {S_2} = \dfrac{n}{2}\left[ {4n + 6} \right]$,
$ \Rightarrow {S_2} = n\left[ {2n + 3} \right]$,
$ \Rightarrow {S_2} = 2{n^2} + 3n$,
Now considering${S_3}$and again using the formula we get,
Now using the formula of the sum of n terms of the arithmetic progression, A.P i.e., ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$, here a=5, d=6,
Now substituting the values we get,
$\Rightarrow$${S_3} = \dfrac{n}{2}\left[ {2\left( 5 \right) + \left( {n - 1} \right)6} \right]$,
Now simplifying we get,
$\Rightarrow$${S_3} = \dfrac{n}{2}\left[ {10 + 6n - 6} \right]$,
$ \Rightarrow {S_3} = \dfrac{n}{2}\left[ {6n + 4} \right]$,
$ \Rightarrow {S_3} = n\left[ {3n + 2} \right]$,
$ \Rightarrow {S_3} = 3{n^2} + 2n$,
Now we have to prove that${S_1} + {S_3} = 2{S_2}$, now substituting the values we get,
$ \Rightarrow {n^2} + 4n + 3{n^2} + 2n = 2\left( {2{n^2} + 3n} \right)$
Now simplifying we get,
$\Rightarrow$$4{n^2} + 6n = 4{n^2} + 6n$,
Hence proved.

The sum of first n terms of three A.P s are ${S_1},{S_2},{S_3}$ . The first term of each is 5 and their common differences are 2, 4, and 6 respectively, then ${S_1} + {S_3} = 2{S_2}$.

Note:
In these type of questions students should know the formulas of the arithmetic progression and the sum of n terms and the nth term, and the sum of n terms formula is given by \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\] and nth term formula is given by ${T_n} = 2a + \left( {n - 1} \right)d$, where a is the first term, d is the common difference.