Question

How do you find the nth term of the sequence $ \dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},\dfrac{4}{5},.......? $

Answer 1

Hint: We will first recall the concept of the sequence and series and then we will analyze the given sequence and find the general term of the sequence. Since we can see that numerator occur at the difference of 1 and the first term is 1 so, we can say that numerator is in Arithmetic Progression and the denominator also occur at the difference of 1 and its first term is 2 so, we can say that denominator is in Arithmetic Progression. So, we will use the $ nth $ term formula $ {{T}_{n}}=a+\left( n-1 \right)d $ , where a is the first term and d is a common difference.

Complete step by step answer:
We know that the sequence is the list of the number or object in a special order.
We can see that each term is a proper fraction. So, we can first find the nth term of the sequence for the numerator and denominator independently, and then we will divide the general terms of numerator and denominator to obtain the nth term of the given sequence.
 And, after seeing the sequence we can analyze that numerator occurs at the difference of 1 and the first term of the sequence is 1. Since the numerator occurs at the interval of 1 so, we can say that the numerator is in Arithmetic Progression.
So, the nth term of the Arithmetic Progression whose first term is ‘a’ and the common difference is ‘d’ is given as $ {{T}_{n}}=a+\left( n-1 \right)d $
So, the nth term for the numerator whose first term is 1 and the common difference is 1 is given as:
 $ \begin{align}
  & \Rightarrow {{T}_{n}}=1+\left( n-1 \right)\times 1 \\
 & \Rightarrow {{T}_{n}}=n-1+1 \\
 & \Rightarrow {{T}_{n}}=n \\
\end{align} $
So, the nth term of the numerator of the given sequence is $ n $ .
Now, after seeing the denominator of the given sequence we can see that the first term of the sequence is 2 and the denominator occurs at the difference is 1. Since the numerator occur at the interval of 1 so, we can say that the numerator is in Arithmetic Progression.
So, the nth term for the numerator whose first term is 2 and the common difference is 1 is given as:
 $ \begin{align}
  & \Rightarrow {{T}_{n}}=2+\left( n-1 \right)\times 1 \\
 & \Rightarrow {{T}_{n}}=n-1+2 \\
 & \Rightarrow {{T}_{n}}=n+1 \\
\end{align} $
So, the nth term of the denominator of the given sequence is $ \left( n+1 \right) $ .
Hence, we can write the nth term of the sequence $ \dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},\dfrac{4}{5},....... $ by dividing the nth term of the numerator and denominator.
So, the nth term of the sequence $ \dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},\dfrac{4}{5},....... $ is $ \dfrac{\text{nth term of the numerator}}{\text{nth term of the denominator}}=\dfrac{n}{n+1} $
Now, we will verify our result by finding any one term of the sequence.
When we put n = 2 in $ \dfrac{n}{n+1} $ , we will get:
 $ \begin{align}
  & =\dfrac{2}{2+1} \\
 & =\dfrac{2}{3} \\
\end{align} $
Since, we can see from the sequence $ \dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},\dfrac{4}{5},....... $ the $ \dfrac{2}{3} $ is the $ {{2}^{nd}} $ term which is correct, so this verifies our result.


Note:
 We must verify our nth term or the general term of the sequence which we find for the given sequence by finding one or more terms of the sequence so that we didn’t get the wrong result. We will verify our result by finding any one term of the above sequence.
Now, we put n = 2 in $ \dfrac{n}{n+1} $ , we will get:
 $ \begin{align}
  & =\dfrac{2}{2+1} \\
 & =\dfrac{2}{3} \\
\end{align} $
Since, we can see from the sequence $ \dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},\dfrac{4}{5},....... $ the $ \dfrac{2}{3} $ is the $ {{2}^{nd}} $ term which is correct, so this verifies our result.