Question

please answer it fast I will market as brainist this is the question the sum of first 6 terms of an ap is 42. The ratio of its 10th term to its 30th term is 1 : 3 calculate the first term and 13th term​

Answer 1

✔✔✅ Answer ✅✔✔

S6 =42 a + 9d 1 ------------ = ------- a + 29d 3 cross multiply we get 3a + 27d = a +29 d 2a - 2d = 0 ------------------ (1) its given that sum of first six terms of an AP is 42 therefore S6 = n/2 ( 2a + (n-1) d) 42 = 6/2 ( 2a + (6-1) d) 42 = 3 (2a + 5d ) 14 = 2a +5d 2a +5d = 14 ------------------- (2) solve eq 1 and 2 2a - 2d = 0 2a +5d = 14 we get d= 2 a = 2 --------------- 13th term of AP = a + (n-1)d 2+ (13-1) 2 = 2+ 24 =26