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How do you write the rule or the $n^{th}$ term given $\dfrac{1}{4},\dfrac{2}{5},\dfrac{3}{6},\dfrac{4}{7},\dfrac{5}{8},...$ ?

Answer
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Hint: In this question, we have to find the $n^{th}$ term of a given sequence. Thus, we will use the arithmetic sequence rule to get the solution. As we see that both the numerator and the denominator the given fractional are in the form of arithmetic sequence, because both have the common difference equal to 1. Also, we see that the denominator is equal to the sum of numerator and 3. Thus, for the $n^{th}$ term, we will let the numerator equal to n and then find the denominator in terms of n, to get the required solution for the problem.

Complete step by step answer:
According to the problem, we have to find the $n^{th}$ term of the sequence.
Thus, we will use the arithmetic sequence rule to get the solution.
The sequence given to us is $\dfrac{1}{4},\dfrac{2}{5},\dfrac{3}{6},\dfrac{4}{7},\dfrac{5}{8},...$ -------- (1)
Now, we see that the numerator of all the terms in equation (1) are in the form of an arithmetic sequence, that is the terms are $1,2,3,4,5,...$ . Therefore the common difference in any two consecutive terms is equal to 1.
Similarly, the denominator of all the terms in equation (1) is $4,5,6,7,8,...$ , thus the common difference between any two consecutive terms is equal to 1, which implies it is an arithmetic sequence.
Now, for the $n^{th}$ term, let us suppose the numerator is equal to n. Also, the denominator is greater than the numerator by 3, which implies the denominator is equal to the sum of numerator and 3, thus we get
$\Rightarrow \text{denominator=numerator+3}$
Now, numerator is equal to n, therefore the above equation will become
$\Rightarrow \text{denominator}=n+3$

Therefore, the $n^{th}$ term will become $\dfrac{n}{n+3}$ for the sequence $\dfrac{1}{4},\dfrac{2}{5},\dfrac{3}{6},\dfrac{4}{7},\dfrac{5}{8},...$.

Note: While solving this problem, do mention all the steps properly to avoid error and confusion. For checking this problem, let n=1, 2, 3, and so on to get the required sequence of the solution.