Question

What is the nth term of the sequence 1,5,9,13,17…?
A.2n-1
B.2n+1
C.4n-3
D.None of the above

Answer 1

Hint: In this question, we are given the first 5 numbers of the series and we have to find the next number. Therefore, we should try to find the relation between consecutive numbers, which in this case will give us that the series is an arithmetic progression. Then using the formula for finding the nth term of an AP, ${{a}_{n}}={{a}_{0}}+(n-1)d$ where ${{a}_{0}}$ is the first term and d is the common difference, we can find the next term in the series.

Complete step-by-step answer:
We know that in an arithmetic progression, the next term in a series is obtained by adding a fixed number d to the previous term and d is known as the common difference……………………………………(1.1)
In this question, the first five terms of the series are given as 3, 6, 9, 12, 15. If we take the difference of the consecutive terms, i.e. subtracting the previous term from a term, we get
$\begin{align}
  & 5-1=4 \\
 & 9-5=4 \\
 & 13-9=4 \\
 & 17-13=4 \\
\end{align}$
Therefore, we find that each successive term is obtained by adding 4 to the previous term. Therefore, comparing this to (1.1), we find that this series is in an arithmetic progression with common difference 4 and first term 1……………………..(1.2)
We know that the formula for finding the nth term of an arithmetic progression with first term ${{a}_{0}}$ and common difference d is given by
${{a}_{n}}={{a}_{0}}+(n-1)d......................(1.3)$
Therefore, from (1.2), taking ${{a}_{0}}=1$ and $d=4$ and using it in (1.3), we find that the nth term of the given series should be
${{a}_{n}}={{a}_{0}}+(n-1)d=1+\left( n-1 \right)\times 4=1+4n-4=4n-3$
Thus the answer is 4n-3 which matches option (c) of the question. Therefore, (c) should be the correct answer to this question.

Note: As we found that the series is in an arithmetic progression with first term ${{a}_{0}}=1$ and common difference $d=4$ , we should note that we should take (n-1) in equation (1.3) and not n as the difference gets added from the second term, i.e. in the first term, 0 times d is added, in the second term 1 times d is added and so on.