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

Lets first calculate $$54^{th}$$ of the given AP.

First term = a = 3,
Common difference = d
= 15 – 3 = 12

Using formula $$a_n = a + (n - 1)d$$, to find nth term of arithmetic progression,

$$a_{54}$$ \)
$$= a + (54 - 1) d$$
$$= 3 + 53 (12)$$
= 3 + 636 = 639

We want to find which term is 132 more than its $$54^{th}$$ term.

Let us suppose it is nth term which is 132 more than $$54^{th}$$ term.

$$\Rightarrow a_n = a_{54} + 132$$
$$\Rightarrow$$ 3 + (n - 1) 12 = 639 + 132
$$\Rightarrow$$ 3 + 12n – 12 = 771
$$\Rightarrow$$ 12n – 9 = 771
$$\Rightarrow$$12n = 780
$$\Rightarrow$$ n = 65

Therefore, 65th term is 132 more than its $$54^{th}$$ term.