Multiple choice

The minimum number of terms of the series $1+3+9+27+.....$ so that the sum may exceed $1000$, is

  1. $7$
  2. $5$
  3. $3$
  4. $None\ of\ these$
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A Correct answer
Explanation

This is a geometric series with a = 1 and r = 3. Sum = a(r^n - 1) / (r - 1) = (3^n - 1) / 2. We want (3^n - 1) / 2 > 1000, so 3^n - 1 > 2000, 3^n > 2001. 3^6 = 729, 3^7 = 2187. Thus, n must be at least 7.

AI explanation

The given sequence is a geometric progression with the first term a = 1 and common ratio r = 3. The sum of the first n terms of a geometric progression is given by the formula S_n = a(r^n - 1) / (r - 1), so here S_n = (3^n - 1) / 2. Checking n = 7 gives S_7 = (3^7 - 1) / 2 = (2187 - 1) / 2 = 1093, which is the first sum to exceed 1000. Therefore, the minimum number of terms required is 7.

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