Question

The $4^{\text {th }}$ term of $G P$ is $500$ and its common ratio is $\frac{1}{m}, m \in N$ Let $S_n$ denote the sum of the first $n$ terms of this GP If $S_6>S_5+1$ and $S_7< S_6+\frac{1}{2}$, then the number of possible values of $m$ is

Answer 1

Correct Answer: 12

Given a geometric progression (GP), the formula for the nth term is \(a_n = ar^{n-1}\). The 4th term is 500, so \(a \cdot \left(\frac{1}{m}\right)^{3} = 500\). Thus, \(a = 500m^3\).

The sum of the first n terms of a GP, when the common ratio \(r \neq 1\), is \(S_n = a \frac{r^n - 1}{r - 1}\).

The 4th term of the GP is given by: \[ T_4 = ar^3 = 500, \] where \( a \) is the first term and \( r = \frac{1}{m} \) is the common ratio. Thus: \[ a \cdot \left(\frac{1}{m}\right)^3 = 500 \implies a = 500 \cdot m^3. \] The sum of the first \( n \) terms of the GP is: \[ S_n = a \frac{1 - r^n}{1 - r}, \quad \text{where } r = \frac{1}{m}. \] Using the conditions: 1. \( S_6 > S_5 + 1 \), 2. \( S_7 < S_6 + \frac{1}{2} \). 

Condition 1: \( S_6 > S_5 + 1 \): \[ S_6 - S_5 > 1 \implies ar^5 > 1. \] Substitute \( a = 500 \cdot m^3 \) and \( r = \frac{1}{m} \): \[ 500 \cdot m^3 \cdot \left(\frac{1}{m}\right)^5 > 1 \implies \frac{500}{m^2} > 1 \implies m^2 < 500. \] 

Condition 2: \( S_7 < S_6 + \frac{1}{2} \): \[ S_7 - S_6 < \frac{1}{2} \implies ar^6 < \frac{1}{2}. \] Substitute \( a = 500 \cdot m^3 \) and \( r = \frac{1}{m} \): \[ 500 \cdot m^3 \cdot \left(\frac{1}{m}\right)^6 < \frac{1}{2} \implies \frac{500}{m^3} < \frac{1}{2} \implies m^3 > 1000. \] 

Combining conditions: \[ m^2 < 500 \quad \text{and} \quad m^3 > 1000. \] The values of \( m \) that satisfy both conditions are: \[ m = 11, 12, 13, \dots, 22. \] The total number of possible values of \( m \) is: \[ \boxed{12}. \]