Multiple choice

Which term of the given sequence will be 132 more than its 58th term? 3, 15, 27, 39, ...

  1. 63

  2. 65

  3. 67

  4. 69
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D Correct answer
Explanation

The sequence is an arithmetic progression with first term a = 3 and common difference d = 12. The 58th term is a + 57d = 3 + 57 * 12 = 3 + 684 = 687. We want the term that is 132 more than this, so 687 + 132 = 819. Setting a + (n-1)d = 819, we get 3 + (n-1)12 = 819, so (n-1)12 = 816, n-1 = 68, n = 69.

AI explanation

The given sequence is an arithmetic progression with the first term a = 3 and the common difference d = 12. Using the formula for the nth term, T_n = a + (n - 1)d, the 58th term is 3 + 57 times 12, which equals 687. We need the term that equals 132 more than 687, which is 819; solving 3 + (n - 1) times 12 = 819 gives n - 1 = 68, so n = 69.

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