Question

the sum of first six terms of an AP is 42.The ratio of the 10th term to its 30th term is 1:3.calculate the first term and 13 th term of the AP?

Answer 1

let first term of A.P = a

let common difference of A.P = d

given that sum of first six term of A.P = 42

Sum of series = Sn = n/2( 2a + (n-1)d )

⇒ S6 = 6/2( 2a + (6-1)d ) = 42

⇒  6a + 15d = 42

⇒  2a + 5d = 14   -----------------------equation 1

tenth term a10 = a + ( 10 - 1 ) d = a + 9d

30th term a30 = a + ( 30 - 1) d = a + 29d

givne that a10/a30  = (a + 9d)/(a + 29d) = 1/3

⇒3(a + 9d) = a + 29d

⇒3a + 27d = a + 29d

⇒2a - 2d = 0

⇒ a = d

substituting a as d in eq 1

2d + 5d = 14

⇒ 7d = 14

⇒ d = 2

⇒ a = d = 2

13th term = a13 = a + (13-1)d

⇒ 2 + (12) × 2 = 26

hence first term is 2 and 13th term is 26.