Question

If the ${{n}^{th}}$ term of the A.P is $\left( 2n-1 \right)$, then the sum of its first $n$ terms will be:
A. ${{n}^{2}}-1$
B. ${{\left( 2n-1 \right)}^{2}}$
C. ${{n}^{2}}$
D. ${{n}^{2}}+1$

Answer 1

Hint: A.P is a sequence in which every set of consecutive terms have a common difference. We will find the first three terms by putting n = 1, 2 and 3 in the ${{n}^{th}}$ term of the arithmetic sequence, which is 2n - 1. Then we will add those terms to find the sum of the first three terms and then repeat it for $n$ terms.

Complete step by step answer:
According to the question, it is asked to find the sum of its first $n$ terms if the ${{n}^{th}}$ term of the A.P is $\left( 2n-1 \right)$. So, as we know that the arithmetic progression (A.P) has affixed common difference $d$ within any two continuous terms of this series and the formula of A.P is given as:
${{a}_{n}}=a+\left( n-1 \right)d$
Here ${{a}_{n}}$ is the ${{n}^{th}}$ term, $a$ is the first term and $n$ is the number of that term which is to be found. So, as here it is given that the last term or ${{a}_{n}}$ is given as $\left( 2n-1 \right)$. So, if we find the first some terms, then,
${{a}_{n}}=a+\left( n-1 \right)d=2n-1$
For $n=1$,
$\begin{align}
  & {{a}_{1}}=a+\left( 1-1 \right)d=2\left( 1 \right)-1 \\
 & {{a}_{1}}=a=1 \\
\end{align}$
For $n=2$,
${{a}_{2}}=2\left( 2 \right)-1=4-1=3$
So, the common difference $d={{a}_{2}}-{{a}_{1}}=3-1=2$.
Sum of $n$ terms is,
$\begin{align}
  & =\left( \dfrac{n}{2} \right)\left( 2a+\left( n-1 \right)d \right) \\
 & =\left( \dfrac{n}{2} \right)\left( 2+\left( n-1 \right)2 \right) \\
 & =\left( \dfrac{n}{2} \right)\left( 2+2n-2 \right) \\
 & =\left( \dfrac{n}{2} \right)\left( 2n \right)={{n}^{2}} \\
\end{align}$

So, the correct answer is “Option C”.

Note: We can also do these types of questions if they ask for some digits or not to the infinite term by finding the first, second or all the terms and adding them or by dividing with the help of the formula of sum of all terms. And find the values of any term by ${{a}_{n}}=a+\left( n-1 \right)d$.