The sequence a_ n-1 , n N is an arithmetical progression and d is its common difference. If Lim _ n (1- d^2 a₁^2 ) (1- d^2 a₂^2 ) (1- d^2 a_n^2 ) converges to 1 4 and a₁=8 , then find the value of d

Step-by-step solution

aligned & _ k=1 ^n (1- d^2 a_k^2 )= _ k=1 ^n a_k^2-d^2 a_k^2 = _ k=1 ^n (a_k-d ) a_k (a_k+d ) a_k = _ k=1 ^n a_ k-1 a_k a_ k+1 a_k [ array l a_ k-d =a_ k-1 a_ k+d =a_ k+1 array ] & .= ( a₀ a₁ a₁ a₂ a_ n-1 a_n ) ( a₂ a₁ a₃ a₂ a_ n+1 a_n )= ( a₀ a_n ) ( a_ n+1 a₁ )= ( a₀ a₁ ) ( a_ n+1 a_n ) [but a₀=a₁-d ] & = ( a₁-d a₁ ) ( a_n+d a_n )= (1- d a₁ )(1+ d a₁+(n-1) d _ zeroas n )=1- d a₁ & but 1- d a ₁ = 1 4 3 4 = d a ₁ 8 3 4 = d aligned Hence d=6 .