Question

In an A.P, if the sum of n terms is ${{S}_{n}}=n\left( 4n+1 \right)$ , then find the general form of an A.P.

Answer 1

Hint: For solving this question first we will understand what we mean by arithmetic progression and then we will see the formula for the ${{n}^{th}}$ term of an arithmetic progression is $a+\left( n-1 \right)d$ where $a$ is the first term and $d$ is the common difference. After that, we will substitute $n$ by $n-1$ in ${{S}_{n}}=n\left( 4n+1 \right)$ to get the expression for the sum of ${{S}_{n-1}}$ . Then, we will use the formula ${{a}_{n}}={{S}_{n}}-{{S}_{n-1}}$ to get the expression of ${{n}^{th}}$ term of the given A.P.

Complete step-by-step solution -
Given:
It is given that, there is an arithmetic progression such that the sum of its first $n$ terms is ${{S}_{n}}=n\left( 4n+1 \right)$ and we have to find the general expression of ${{n}^{th}}$ term of the given arithmetic progression.
Now, first, we will understand when a sequence is called an A.P. and what are important conditions for a sequence to be in arithmetic progression.
Arithmetic Progression:
In a sequence when the difference between any two consecutive terms is equal throughout the series then, such sequence will be called to be in arithmetic progression and the difference between consecutive terms is called as the common difference of the arithmetic progression. If ${{a}_{1}}$ is the first term of an A.P. and common difference of the A.P. is $d$ then, ${{n}^{th}}$ the term of the A.P. can be written as ${{a}_{n}}={{a}_{1}}+d\left( n-1 \right)$ and sum of its first $n$ terms is ${{S}_{n}}=\dfrac{n}{2}\left[ 2{{a}_{1}}+\left( n-1 \right)d \right]$. Moreover, for any series if ${{S}_{n}}$, ${{S}_{n-1}}$ represents the sum of its first $n$, $n-1$ terms respectively, then, it is obvious that ${{n}^{th}}$ term of such series will be ${{a}_{n}}={{S}_{n}}-{{S}_{n-1}}$.
Now, we will use the formula ${{a}_{n}}={{S}_{n}}-{{S}_{n-1}}$ to write the expression for ${{n}^{th}}$ term of the given arithmetic progression.
Now, for ${{S}_{n-1}}$ we will substitute $n$ by $n-1$ in ${{S}_{n}}=n\left( 4n+1 \right)$ . Then,
$\begin{aligned} & {{S}_{n}}=n\left( 4n+1 \right) \\\ & \Rightarrow {{S}_{n-1}}=\left( n-1 \right)\left( 4\left( n-1 \right)+1 \right) \\\ & \Rightarrow {{S}_{n-1}}=\left( n-1 \right)\left( 4n-4+1 \right) \\\ & \Rightarrow {{S}_{n-1}}=\left( n-1 \right)\left( 4n-3 \right) \\\ & \Rightarrow {{S}_{n-1}}=4{{n}^{2}}-3n-4n+3 \\\ & \Rightarrow {{S}_{n-1}}=4{{n}^{2}}-7n+3 \\\ \end{aligned}$
Now, we will substitute ${{S}_{n-1}}=4{{n}^{2}}-7n+3$ and ${{S}_{n}}=n\left( 4n+1 \right)$ in the formula ${{a}_{n}}={{S}_{n}}-{{S}_{n-1}}$ . Then,
$\begin{aligned} & {{a}_{n}}={{S}_{n}}-{{S}_{n-1}} \\\ & \Rightarrow {{a}_{n}}=n\left( 4n+1 \right)-\left( 4{{n}^{2}}-7n+3 \right) \\\ & \Rightarrow {{a}_{n}}=4{{n}^{2}}+n-4{{n}^{2}}+7n-3 \\\ & \Rightarrow {{a}_{n}}=8n-3 \\\ \end{aligned}$
Now, from the above result, we conclude that ${{n}^{th}}$ term of the given A.P will be ${{a}_{n}}=8n-3$ . Then,
$\begin{aligned} & {{a}_{n}}=8n-3 \\\ & \Rightarrow {{a}_{1}}=8-3=5 \\\ & \Rightarrow {{a}_{2}}=16-3=13 \\\ \end{aligned}$
Now, from the above result, we can say that, first and second term of the A.P will be $5$ , $13$ respectively. Moreover, as the coefficient of $n$ in ${{a}_{n}}=8n-3$ is $8$ so, the common difference will be $8$.
Thus, we can say that given A.P will be $5,13,21,29,37...................,8n-3$.

Note: Here, the student should know the concept of A.P. and how to express the general expression of the ${{n}^{th}}$ term of an A.P and formula for the sum of its first $n$ terms and important points should be remembered so that question can be answered quickly and correctly without any confusion. Moreover, we could have solved the question by putting $n=1$ in ${{S}_{n}}=n\left( 4n+1 \right)$ to get the value of ${{a}_{1}}=5$ and then $n=2$ in ${{S}_{n}}=n\left( 4n+1 \right)$ to get the value of ${{a}_{1}}+{{a}_{2}}=18$ . Then, we will subtract both results to get ${{a}_{2}}=13$ and further, we can get the value of common difference by the formula $d={{a}_{2}}-{{a}_{1}}$ . After that, we can use the formula ${{a}_{n}}={{a}_{1}}+d\left( n-1 \right)$ to write the general form of the given A.P.