Let a , 1 a 2 , … , a n be a given A.P. whose common difference is an integer and S n = a 1 + a 2 + … + a n . If a 1 = 1 , a n = 300 and 15 ≤ n ≤ 50 , then the ordered pair S n - 4 , a n - 4 is equal to:

Step-by-step solution

a n = a 1 + ( n - 1 ) d 300 = 1 + ( n - 1 ) d ⇒ d = 299 ( n - 1 ) = 13 × 23 ( n - 1 ) = integer so n - 1 = ± 13 , ± 23 , ± 299 , ± 1 ⇒    n = 14 , - 12 , 24 , - 22 , 300 , - 298 , 2 , 0 But n ∈ [ 15 ,   50 ]   ⇒ n = 24   ⇒ d = 13 Hence, S n - 4 = S 20 = 20 2 2 ( 1 ) + ( 20 - 1 ) ( 13 ) ⇒ S n - 4 = 2490 And, a n - 4 = a 20 = a 1 + 19 d = 1 + 19 × 13 = 248