Question

The sum of first term of an ap is 42 the ratio of 10 term to its 30 term is 1:3 find 1 and 13 terms

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