Let a₁, a₂, a₃, be an arithmetic progression with a₁ = 5 . If _ k=1 ²⁰ 1 a_k a_ k+1 = 4 45 , then the value of a₁₁ is equal to _____ .

Step-by-step solution

Let d be the common difference of the arithmetic progression. The general term of the given series can be rewritten as: 1 a_k a_ k+1 = 1 d ( a_ k+1 - a_k a_k a_ k+1 ) = 1 d ( 1 a_k - 1 a_ k+1 ) Summing this from k = 1 to 20 , we get a telescoping series: _ k=1 ²⁰ 1 a_k a_ k+1 = 1 d ( 1 a₁ - 1 a₂ + 1 a₂ - 1 a₃ + + 1 a₂₀ - 1 a₂₁ ) _ k=1 ²⁰ 1 a_k a_ k+1 = 1 d ( 1 a₁ - 1 a₂₁ ) 1 d ( a₂₁ - a₁ a₁ a₂₁ ) = 1 d ( 20d a₁ a₂₁ ) = 20 a₁ a₂₁ We are given that this sum is equal to 4 45 and a₁ = 5 . Substituting these values: 20