Question

Which term of the AP: 4, 9, 14,........, is 89?
A. 18
B. 14
C. 12
D. 20

Answer 1

Hint: To find the term of the AP 4, 9, 14,........ that is 89, we will use the formula ${{a}_{n}}=a+\left( n-1 \right)d...(i)$ , where a is the first term of the AP, d is the common difference that is given by $d={{a}_{2}}-{{a}_{1}}=\text{Second term}-\text{first term}$ and ${{a}_{n}}$ is the ${{n}^{th}}$ term. Here, we will take ${{a}_{n}}=89$ . Now, substitute the values in formula (i) and find the value of n which is the required answer.

Complete step-by-step solution:
We have to find the term of the AP 4, 9, 14,........ that is 89.
We know that the first term of the given AP is 4. We can denote this as
$a=4$
Now, we can now find the common difference, d. This is given by
$d={{a}_{2}}-{{a}_{1}}=\text{Second term}-\text{first term}$
From the given AP, the second term is 9. Hence,
$d=9-4=5$
We know that ${{n}^{th}}$ term of an AP is given by the formula
${{a}_{n}}=a+\left( n-1 \right)d...(i)$
Let us take ${{a}_{n}}=89$ . We have to find n, that is, the term of 89.
Now, let’s substitute the values in equation (i), we will get
$89=4+\left( n-1 \right)\times 5$
Let us solve the RHS. We will get
$\begin{align}
  & 89=4+5n-5 \\
 & \Rightarrow 89=5n-1 \\
\end{align}$
Let us collect constant terms to one side. This is shown below.
$5n=89+1$
On adding the numbers in RHS, we will get
$5n=90$
To find n, we have to take 5 to RHS. This is shown below.
$n=\dfrac{90}{5}$
On simplifying this, we will get
$n=18$
Hence ${{18}^{th}}$ term of the AP 4, 9, 14,........ is 89.
Hence, the correct option is A.

Note: You may make mistakes when writing the formula for common difference as $d={{a}_{1}}-{{a}_{2}}$ . This will lead to a negative value. There can also be a chance of making error when writing the formula for ${{n}^{th}}$ term of an AP as ${{a}_{n}}=a-\left( n+1 \right)d$ . If students do not remember the formula, they can just observe the pattern of the series and then fill in the terms by adding 5 to the previous term. This will be time-consuming since the options are all above 10 and the maximum is 20, so students will have to find the terms manually. If the options were all less than 10, this method would have worked effectively.