<p>Let \(a_1, a_2, a_3, \ldots, a_{101}\) be in G.P. with \(a_{101} = 25\) and \(\displaystyle\sum_{i=1}^{201} a_i = 625\). Then the value of \(\displaystyle\sum_{i=1}^{201} \frac{1}{a_i}\) equals ___.</p>
Step-by-Step Solution
Key Concept: In a G.P., if the sum of terms equals S and the last term is L, the sum of reciprocals can be found using the reciprocal property: the reciprocals form a G.P. in reverse order with sum = S/L when the G.P. has specific symmetry. Alternatively, use the relationship that Σ(1/aᵢ) = Σ(aₙ₊₁₋ᵢ)/product, leveraging the geometric mean property.
<p><strong>Step 1:</strong> Set up the G.P. with first term 'a' and common ratio 'r'. We have 101 terms: a₁, a₂, ..., a₁₀₁.</p><p>Given: a₁₀₁ = ar¹⁰⁰ = 25</p><p><strong>Step 2:</strong> Sum of G.P. is Σ(i=1 to 101) aᵢ = a(r¹⁰¹ - 1)/(r - 1) = 625</p><p><strong>Step 3:</strong> For the sum of reciprocals: Σ(i=1 to 101) (1/aᵢ) = Σ(i=1 to 101) 1/(ar^(i-1)) = (1/a)·Σ(i=1 to 101) (1/r)^(i-1)</p><p>This is a G.P. with first term (1/a) and common ratio (1/r).</p><p><strong>Step 4:</strong> Σ(1/aᵢ) = (1/a)·[(1/r)¹⁰¹ - 1]/[(1/r) - 1] = (1/a)·[1 - r¹⁰¹]/[r¹⁰⁰(1-r)]</p><p><strong>Step 5:</strong> Simplify: Σ(1/aᵢ) = [1 - r¹⁰¹]/[ar¹⁰⁰(1-r)] = [Σ(aᵢ)]/[ar¹⁰⁰] = 625/25 = 25</p><p>∴ Answer: <strong>25</strong></p>
Correct Answer: 25