Question

An AP has the first term as 2 and the common difference as 4. Which term in the AP is the first multiple of 23 to appear?

• A. ${11}^{th}term$
• B. ${12}^{th}term$
• C. ${14}^{th}term$
• D. ${35}^{th}term$

Answer 1

The correct option is B ${12}^{th}term$

Given a = 2 and d = 4.
$n^{th}term{a}_n=a+(n-1)d$
Any multiple of 23 can be taken as 23k (k = 1, 2, 3, ...).
Given, $a_n=23k$
$a+(n-1)d=23k$
$2+(n-1)4=23k$
$(n-1)4=23k-2$
$n-1=\frac{23k-2}{4}$
Since $n$ must be a natural number, as $n$ is the number of terms of an AP, the expression $\frac{23k-2}{4}$ must be a whole number.
When we put k = 1, $\frac{23k-2}{4}=\frac{23\times 1-2}{4}=\frac{21}{4}$ which is not a whole number.
When we put k = 2, $\frac{23k-2}{4}=\frac{23\times 2-2}{4}=\frac{44}{4}=11$ which is a whole number.
Hence, $23\times 2=46$ is the first multiple of 23 to appear in the AP.
Now, since we have used the formula for finding the $n^{th}$ term, the value of $n$ should give us which term of the AP is 46.
We know that $n-1=11$, hence $n=12$.
So, ${12}^{th}$ term in the AP is the first multiple of 23 to appear in the AP.