Question

nth term of a Sequence Find the nth term of a sequence whose first several terms are given.

$\frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7},$….

Short answer

The nth term of the sequence is $a_n=\frac{n+2}{n+3}$.

Step-by-step solution

Step 1.  Given

The given sequence is $\frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7},$…

Step 2.  To determine

We have to find thenth term of the sequence.

Step 3.  Calculation

The given sequence $\frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7},$…

The numerators are 3, 4, 5, 6…

This is an arithmetic sequence with common difference $d=\text{2nd} \text{term}-1\text{st} \text{term =} 4-3=1$.

First term is $a_1=3$.

Using the term formula for arithmetic sequence we get the nth numerator to be:

$\begin{array}{c}{a}_n=a_1+\left(n-1\right)d\\ =3+\left(n-1\right)\left(1\right)\\ =3+n-1\\ =n+2\end{array}$

The denominators are 4, 5, 6, 7, …

This is an arithmetic sequence with common difference $d=\text{2nd} \text{term}-1\text{st} \text{term =} 5-4=1$.

First term is $a_1=4$.

Using the term formula for arithmetic sequence we get the nth numerator to be:

$\begin{array}{c}{a}_n=a_1+\left(n-1\right)d\\ =4+\left(n-1\right)\left(1\right)\\ =4+n-1\\ =n+3\end{array}$

Hence, the nth term of the sequence is$a_n=\frac{n+2}{n+3}$.