Multiple choice

Starting from the first term, how many terms of the A.P. series: 3, 6, 9, 12, 15, ... must be taken to make the sum 108?

  1. 6

  2. 7

  3. 8

  4. 36
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C Correct answer
Explanation

The series is an AP with a=3 and d=3. The sum formula is S = n/2 * (2a + (n-1)d). Setting 108 = n/2 * (6 + (n-1)3) leads to 108 = n/2 * (3n + 3), which simplifies to 3n^2 + 3n - 216 = 0, or n^2 + n - 72 = 0. Factoring gives (n+9)(n-8) = 0, so n=8.

AI explanation

The arithmetic progression has a first term of 3 and a common difference of 3. Using the sum formula for an arithmetic progression, S = n/2 * [2a + (n - 1)d], we substitute the known values to get 108 = n/2 * [6 + (n - 1)3]. This simplifies to 216 = n(3n + 3), and then to 3n^2 + 3n - 216 = 0. Dividing by 3 gives n^2 + n - 72 = 0, which factors into (n - 8)(n + 9) = 0. Since the number of terms must be positive, n is 8.

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