If a i i = 1 n , where n is an even integer, is an arithmetic progression with common difference 1 , and ∑ i = 1 n a i = 192 , ∑ i = 1 n 2 a 2 i = 120 , then n is equal to

Step-by-step solution

Given, ∑ i = 1 n a i = 192 a 1 + a 2           ⋯   a n = 192 ⇒     n 2 a 1 + a n = 192 a 1 + a n = 384 n           ⋯ 1 Also given ∑ i = 1 n 2 a 2 i = 120 ⇒     a 2 + a 4 + a 6   ⋯ a n   ⏟ n 2   terms = 120 ⇒     n 2 × 1 2 a 2 + a n = 120 a 2 + a n = 480 n a 1 + 1 + a n = 480 n       ⋯ 2 Now equation 2 - equation 1 480 n - 384 n = a 1 + a