#### Question

Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54^{th}term?

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#### Solution

Given A.P. is 3, 15, 27, 39, …

*a *= 3

*d* = *a*_{2} − *a*_{1} = 15 − 3 = 12

*a*_{54} = *a* + (54 − 1) *d*

= 3 + (53) (12)

= 3 + 636 = 639

132 + 639 = 771

We have to find the term of this A.P. which is 771.

Let *n*^{th} term be 771.

*a*_{n} = *a* + (*n* − 1) *d*

771 = 3 + (*n* − 1) 12

768 = (*n* − 1) 12

(*n* − 1) = 64

*n* = 65

Therefore, 65^{th} term was 132 more than 54^{th} term.

**Alternatively,**

Let *n*^{th} term be 132 more than 54^{th} term.

`n = 54 + 132/12`

= 54 + 11 = 65^{th} term

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#### Reference Material

Solution for question: Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54th term? concept: nth Term of an AP. For the course CBSE