Multiple choice

Find the least value of n for which the series $3+6+9+....\dots$ upto $n$ terms exceeds $1000$.

  1. $20$
  2. $24$
  3. $25$
  4. $26$
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D Correct answer
Explanation

The sum of the arithmetic series is n/2 * (2*3 + (n-1)*3) = 3n(n+1)/2. We need 1.5n(n+1) > 1000, so n(n+1) > 666.67. Testing values, 25*26 = 650 (too small) and 26*27 = 702 (exceeds 1000).

AI explanation

The given series is an arithmetic progression with the first term a equal to 3 and the common difference d equal to 3. Using the sum formula for an arithmetic progression, S equals n divided by 2 times the quantity 2a plus the quantity n minus 1 times d, we substitute the values to get the equation 1000 equals n divided by 2 times the quantity 6 plus the quantity n minus 1 times 3. Solving the resulting quadratic equation 3n squared plus 3n minus 2000 equals 0 gives a positive root of approximately 25.4, which means the sum first exceeds 1000 at n equals 26.

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