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Which term of AP $3, 15, 27, 39, . .$ will be $132$ more than its $54^{th}$ term?

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a65

a64

a60

None of these

answer is A.

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Detailed Solution

Given, the sequence is

3, 15, 27, 39, \dots

Common difference, d=15-3

\Rightarrow d = 12

a = 3

Let the

n^{th}

term be the

132

more than

54^{th}

term of the given AP.
nth term of an AP is calculated using

a_{n} = a + (n - 1) d .

Substitute values,

\Rightarrow a_{n} = a_{54} + 132

\Rightarrow a + (n - 1) d = a + 53 d + 132

\Rightarrow (n - 1) \times 12 = 53 \times 12 + 132

\Rightarrow n - 1 = 53 + 11

\Rightarrow n = 65

Hence,

a_{65}

132

more than

a_{54}

.
Therefore, option 1 is correct.

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