Question

If the series 3+7+11+15+............ is given then find ${17}^{th}$ term of the series.
(a) 63
(b) 65
(c) 67
(d) 69

Answer 1

Hint: For solving this question we will first analyse the given series and we will try to give the general expression for the ${{n}^{th}}$ term of the given series. Then, we will find the ${17}^{th}$ term of the given series and select the correct option.

Complete step-by-step answer:
Given:
We have to find the ${17}^{th}$ term of the series 3 + 7 + 11 + 15 + ………. .
Now, the series can be written as $3,5,7,11.........$ . If we analyse the given series then we will find that the difference between any two consecutive terms is the same throughout and that is 4. For example: If ${{T}_{r}}$ represents the ${{r}^{th}}$ term of the given series then, ${{T}_{2}}-{{T}_{1}}=7-3=4$ , ${{T}_{3}}-{{T}_{2}}=11-7=4$ and ${{T}_{4}}-{{T}_{3}}=15-11=4$ .
Now, such series in which the difference between consecutive terms is equal are called Arithmetic progression (A.P.). And if ${{a}_{1}}$ is the first term of the A.P. and $d$ is the common difference.
 Then, ${{n}^{th}}$ term of the A.P. $={{T}_{n}}={{a}_{1}}+\left( n-1 \right)d$ .
Now, we will use the above result directly to get the general expression of the ${{n}^{th}}$ term for the given series. We have to put ${{a}_{1}}=3$ and $d=4$ in the above expression of ${{T}_{n}}$ . Then,
$\begin{align}
  & {{T}_{n}}={{a}_{1}}+\left( n-1 \right)d \\
 & \Rightarrow {{T}_{n}}=3+\left( n-1 \right)4 \\
 & \Rightarrow {{T}_{n}}=3+4n-4 \\
 & \Rightarrow {{T}_{n}}=4n-1...................\left( 1 \right) \\
\end{align}$
Now, equation (1) represents the general expression of the ${{n}^{th}}$ term of the given series. So, we can find the ${17}^{th}$ term of the given series by substituting $n=17$ in the equation (1). Then,
$\begin{align}
  & {{T}_{n}}=4n-1 \\
 & \Rightarrow {{T}_{17}}=4\times 17-1 \\
 & \Rightarrow {{T}_{17}}=67 \\
\end{align}$
Thus, 67 is the ${17}^{th}$ term of the series $3+7+11+15+............$ .
Hence, (c) is the correct option.

Note: Here, as the series, the common difference was the same that’s why we used the concept of Arithmetic progression. We could have solved this by another method also if we analyse the given series and add 1 to each term we will get the multiples of 4 in a proper sequence, so we can check for the correct option using this method.