Let a 1 , a 2 , a 3 , . . . . , a n be n positive consecutive terms of an arithmetic progression. If d > 0 is its common difference, then lim n → ∞ d n 1 a 1 + a 2 + 1 a 2 + a 3 + … + 1 a n - 1 + a n is

Step-by-step solution

Given, a 1 , a 2 , a 3 , . . . . , a n are terms of an A . P , So common difference will be, d = a 2 - a 1 = a 3 - a 2 = . . . . . . . . = a n - a n - 1 Now solving, lim n → ∞ d n 1 a 1 + a 2 + 1 a 2 + a 3 + … + 1 a n - 1 + a n = lim n → ∞ d n a 2 - a 1 a 2 - a 1 + a 3 - a 2 a 3 - a 2 + … + a n - a n - 1 a n - a n - 1 = lim n → ∞ d n × 1 d a n - a 1 Now using the formula a n = a 1 + n - 1 d we get, = lim n → ∞ 1 d a 1 + ( n - 1 ) d - a 1 n = lim n &#8