Question

Find the first five terms of the sequence whose ${{n}^{th}}$ term is ${{t}_{n}}=\dfrac{1}{n+1}$ .

Answer 1

Hint: First, we write down the expression for the ${{n}^{th}}$ term of the sequence. After that, we find the first term by putting the value of n as $1$ , the second term by putting the value of n as $2$ and so on, till we reach the fifth term.

Complete step by step answer:
The general expression for the sequence given in the problem is,
${{t}_{n}}=\dfrac{1}{n+1}$
This is also the expression for the ${{n}^{th}}$ term of the sequence or the ${{n}^{th}}$ term itself. Here, “n” is a variable which represents the position of a term in the sequence. This means that the value of “n” ranges from $1$ to $\infty $ which is the set of all the natural numbers. So, if we want to find the value of any term, we simply need to put the value of n equal to the position of the respective term.
In this problem, we need to find the first five terms of the sequence with the ${{n}^{th}}$ term being ${{t}_{n}}=\dfrac{1}{n+1}$ . For the first term, we need to put the value of $n=1$ and get,
$\Rightarrow {{t}_{1}}=\dfrac{1}{1+1}=\dfrac{1}{2}$
For the second term, we need to put the value of $n=2$ and get,
$\Rightarrow {{t}_{2}}=\dfrac{1}{2+1}=\dfrac{1}{3}$
For the third term, we need to put the value of $n=3$ and get,
$\Rightarrow {{t}_{3}}=\dfrac{1}{3+1}=\dfrac{1}{4}$
For the fourth term, we need to put the value of $n=4$ and get,
$\Rightarrow {{t}_{4}}=\dfrac{1}{4+1}=\dfrac{1}{5}$
For the fifth term, we need to put the value of $n=5$ and get,
$\Rightarrow {{t}_{5}}=\dfrac{1}{5+1}=\dfrac{1}{6}$
Thus, we can conclude that the first five terms are $\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5},\dfrac{1}{6}$ respectively.

Note: At first, we should carefully notice how many terms need to be found out. Also, for this specific problem, we must not mistakenly take the ${{n}^{th}}$ term to be $\dfrac{1}{n}$ as this gives wrong results. Finding out whether the sequence is arithmetic or geometric is irrelevant here.