Question

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

Answer 1

(There is some error in your question.
Ratio given is b/w 10thand 30th term.
Not b/w 10th and 13 th term.i have corrected it solve the question for you.)
Let, a=1st term, d=common difference.

S6 =42
10th term=a+9d
30th term=a+29d
Giben that their ratio=1/3

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