Question

If the sum of first n terms of the series is $\left( n+12 \right)$, then what is its third term?
$\begin{aligned} & \text{A}\text{. 1} \\\ & \text{B}\text{. 2} \\\ & \text{C}\text{. 3} \\\ & \text{D}\text{. 4} \\\ \end{aligned}$

Answer 1

As given the sum of first n terms of the series is $\left( n+12 \right)$, by putting the value of $n=3$ and $n=2$, we calculate the sum of first three terms and sum of first two terms. By subtracting the sum of two terms from the sum of three terms we obtain the third term of the series.

Complete step by step answer:
We have given the sum of the first n terms of the series is $\left( n+12 \right)$.
We have to find the third term of the series.
Now, let us assume ${{S}_{n}}$ denotes the sum of n terms,
So, as given in the question sum of first n terms will be ${{S}_{n}}=\left( n+12 \right)$
Now, to find the third term of the series we have to find the sum of first three terms and sum of first two terms of series then, subtract both the values to get the third term.
Now, to find the sum of first three terms we have to put $n=3$
We get
$\begin{aligned} & {{S}_{3}}=\left( 3+12 \right) \\\ & {{S}_{3}}=15 \\\ \end{aligned}$
Now, to find the sum of first two terms we have to put $n=2$
We get
$\begin{aligned} & {{S}_{2}}=\left( 2+12 \right) \\\ & {{S}_{2}}=14 \\\ \end{aligned}$
Now, the third term is equal to sum of first three terms minus sum of first two terms
Let the third term be denoted by ${{T}_{3}}$ .
So, ${{T}_{3}}={{S}_{3}}-{{S}_{2}}$
${{T}_{3}}=15-14$
${{T}_{3}}=1$
The third term of the series is $1$ .

So, the correct answer is “Option A”.

Note: Here in this question, it is not specified that the given series is an arithmetic series or geometric series, that’s why we will not use any formula for the sum of series. For both arithmetic series and geometric series the formula for sum of n terms is different.