Skip to content

CA Foundation · Quantitative Aptitude · Sequence and Series

A manufacturing unit's production follows the pattern: 1,200 units in Year 1, 1,500 units in Year 2, 1,800 units in Year 3, and so on. If production continues in this arithmetic pattern, in which year will the production first exceed 3,300 units?

Using the arithmetic progression formula aₙ = 1,200 + (n−1)×300, we solve 1,200 + 300(n−1) > 3,300. This gives n > 8, so n = 9. In Year 9, production will be 3,600 units, first exceeding 3,300.

  1. AYear 8
  2. BYear 9
  3. CYear 10Correct
  4. DYear 11

Explanation

This is an arithmetic sequence with a = 1,200, d = 300. The nth term is given by: aₙ = a + (n−1)d = 1,200 + (n−1)×300. We need aₙ > 3,300. Solving: 1,200 + 300(n−1) > 3,300 → 300(n−1) > 2,100 → (n−1) > 7 → n > 8. Thus n = 9. Checking: a₉ = 1,200 + 8(300) = 3,600 > 3,300. Year 8 gives a₈ = 1,200 + 7(300) = 3,300 (not exceeding). Year 9 is the first year production exceeds 3,300 units.

Did you get it right without looking?

One question tells you little. A timed set on Sequence and Series shows your real accuracy, how long you take and where you lose marks.

Practise Sequence and SeriesFree diagnostic mock

More Sequence and Series questions