JEE Main 2021 · 26 February · morning · Single correct · +4 / −1

The sum of the infinite series
1 + 2/3 + 7/3² + 12/3³ + 17/3⁴ + 22/3⁵ + ...... is equal to :

  1. (a) 9/4
  2. (b) 13/4
  3. (c) 15/4
  4. (d) 11/4
Answer and solution

Answer (b) 13/4

Solution

Maths sir · 4 steps

Numerators 2, 7, 12, 17 go up by 5 and denominators grow by 3: an AGP, so subtract S/3 from S.

  1. Spot the pattern. After the first term, the numerators 2, 7, 12, 17, … form an AP with common difference 5, and the denominators are powers of 3. That makes it an arithmetico-geometric series with ratio 1/3.
  2. Shift and subtract. Write S/3 = 1/3 + 2/3² + 7/3³ + … and subtract it from S term by term. Each new numerator is a difference of consecutive numerators: 2 - 1 = 1, then 7 - 2 = 5, 12 - 7 = 5, and so on.

    2S/3 = 1 + 1/3 + 5/3² + 5/3³ + … = 4/3 + (5/9)/(1 - 1/3)

  3. Add the GP. The tail is a GP with first term 5/9 and ratio 1/3, so its sum is 5/9 × 3/2 = 5/6. Then 2S/3 = 4/3 + 5/6 = 13/6.
  4. So. S = 13/6 × 3/2 = 13/4, which is option (b).

The idea to keep. For an AGP with ratio r, compute S - rS. The AP numerators collapse to a constant difference, leaving an ordinary GP you can sum in one line.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Sequences & Series

Asked about 250 times in JEE Main since 2020. This question is from 2021.