The 4ᵗʰ term of GP is 500 and its common ratio is 1/m,m∈ℕ. Let Sₙ denote the sum of the first n terms of this GP. If S₆ > S₅ + 1 and S₇ < S₆ + 1/2, then the number of possible values of m is ___________

Solution

Maths sir · 4 steps

S6 minus S5 is just the 6th term, and S7 minus S6 is the 7th term. Turn both conditions into bounds on m.

1. Find the first term. With r = 1/m, the 4th term is ar³ = a/m³ = 500, so a = 500m³. That gives Tₙ = 500m³ · 1/mⁿ⁻¹.
2. First condition. S₆ - S₅ = T₆ = 500/m². So 500/m² > 1 means m² < 500. Since 22² = 484 and 23² = 529, this gives m ≤ 22.
3. Second condition. S₇ - S₆ = T₇ = 500/m³. So 500/m³ < 1/2 means m³ > 1000, that is m > 10, so m ≥ 11.

   11 ≤ m ≤ 22

4. So. The natural numbers from 11 to 22 all work, which is 22 - 11 + 1 = 12 values of m.

The common slip: 13. Letting m = 10 in, even though m³ > 1000 is strict, gives 13 values instead of 12.

The idea to keep. Whenever a question compares Sₙ₊₁ with Sₙ, rewrite the difference as the single term Tₙ₊₁. Then watch strict versus non-strict signs at the boundary values.

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