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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 in the AP?

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Solution

The correct option is **B** 12th term

Given a = 2 and d = 4

nth term an=a+(n−1)d

Any multiple of 23 can be taken as 23k (k = 1, 2, 3, ...)

Given, an=23k

a+(n−1)d=23k

2+(n−1)4=23k

(n−1)4=23k−2

n−1=23k−24

Since n must be a natural number, as n is the number of terms of an AP, the expression 23k−24 must be a whole number.

When we put k = 1, 23k−24=23×1−24=214 which is not a whole number

When we put k = 2, 23k−24=23×2−24=444=11 which is a whole number

Hence, 23×2=46 is the first multiple of 23 to appear in the AP

Now, since we have used the formula for finding the nth 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, 12th term in the AP is the first multiple of 23 to appear in the AP.

Given a = 2 and d = 4

nth term an=a+(n−1)d

Any multiple of 23 can be taken as 23k (k = 1, 2, 3, ...)

Given, an=23k

a+(n−1)d=23k

2+(n−1)4=23k

(n−1)4=23k−2

n−1=23k−24

Since n must be a natural number, as n is the number of terms of an AP, the expression 23k−24 must be a whole number.

When we put k = 1, 23k−24=23×1−24=214 which is not a whole number

When we put k = 2, 23k−24=23×2−24=444=11 which is a whole number

Hence, 23×2=46 is the first multiple of 23 to appear in the AP

Now, since we have used the formula for finding the nth 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, 12th term in the AP is the first multiple of 23 to appear in the AP.

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