The term of GP is 500 and its common ratio is , . Let denote the sum of the first terms of this GP. If and , then the number of possible values of is
Show answer and solution
Answer: 12
Solution
The 4th term is a·r³ = 500, with r = 1/m, so a = 500m³.
The (n+1)th term equals Sn+1 - Sn, so:
- S6 > S5 + 1 means the 6th term > 1: a·r⁵ = 500m³·(1/m)⁵ = 500/m² > 1, giving m² < 500, so m ≤ 22.
- S7 < S6 + 1/2 means the 7th term < 1/2: a·r⁶ = 500m³·(1/m)⁶ = 500/m³ < 1/2, giving m³ > 1000, so m > 10, meaning m ≥ 11.
So m ∈ {11, 12, 13, ..., 22}, which gives 22 - 11 + 1 = 12 possible values.
The (n+1)th term equals Sn+1 - Sn, so:
- S6 > S5 + 1 means the 6th term > 1: a·r⁵ = 500m³·(1/m)⁵ = 500/m² > 1, giving m² < 500, so m ≤ 22.
- S7 < S6 + 1/2 means the 7th term < 1/2: a·r⁶ = 500m³·(1/m)⁶ = 500/m³ < 1/2, giving m³ > 1000, so m > 10, meaning m ≥ 11.
So m ∈ {11, 12, 13, ..., 22}, which gives 22 - 11 + 1 = 12 possible values.
[JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.)]