Let a₁, a₂, a₃, be an arithmetic progression with positive terms and a common difference d > 0 . If _ n n _ k=1 ^ n 1 a_k + a_ k+2 = 4 , then the value of d is

Step-by-step solution

Let S_n = _ k=1 ^ n 1 a_k + a_ k+2 . Rationalising the general term of the sum: 1 a_k + a_ k+2 = a_ k+2 - a_k a_ k+2 - a_k Since the terms are in an A.P. with common difference d , a_ k+2 - a_k = 2d . Thus, the term becomes: 1 2d ( a_ k+2 - a_k ) Expanding the sum S_n for k=1 to n : S_n = 1 2d [( a₃ - a₁ ) + ( a₄ - a₂ ) + ( a₅ - a₃ ) + + ( a_ n+1 - a_ n-1 ) + ( a_ n+2 - a_n )] This is a telescoping series with a step size of 2. Most terms cancel out, leaving two terms at the beginning and two at the end: S_n = 1 2d