Let P_n denote the product of the first n terms of a geometric progression whose common ratio is a natural number m . If P₅ = 2⁵⁰ , P₈ P₇ > 2²⁵ and P₁₁ P₁₀ < 2⁴² , then the number of possible values of m is

Step-by-step solution

Let the first term of the geometric progression be a and the common ratio be m , where m N . The product of the first 5 terms is given by: P₅ = a (am) (am^2) (am^3) (am^4) = a^5 m¹⁰ Given P₅ = 2⁵⁰ , we have: a^5 m¹⁰ = 2⁵⁰ a m^2 = 2¹⁰ We know that P_n P_ n-1 = T_n , the n -th term of the progression. For the first inequality: P₈ P₇ > 2²⁵ T₈ > 2²⁵ a m^7 > 2²⁵ (a m^2) m^5 > 2²⁵ 2¹⁰ m^5 > 2²⁵ m^5 > 2¹⁵ m > 2^3 m > 8 For the second inequality: P₁₁ P₁₀ a m¹⁰ (a m^2) m^8 2¹⁰ m^8 m^8 m Combining the two conditions, we get