Question

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

Answer 1

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