The sum of first 6 term of an AP is 42 the ratio of its 10th term to its 30th term is 1:3 find 1st & 2nd term

A (n) = a+d (n-1) we are told that a (10) : a (30) is 1:3

(a+9d) / (a+29d) = 1/3

3a+27d=a+29d

2a=2d, a=d ...

s (n) = (a+a+d (n-1)) (n/2)

s (n) = (2a+dn-d) (n/2)

s (n) = (2an+dn^2-dn) / 2 since a=d:

s (n) = (2an+an^2-an) / 2

s (6) = 42 = (12a+36a-6a) / 2

84=42a

a=2 and d=2 so:

a (n) = 2+2 (n-1)

a (n) = 2n

So the first two terms are 2 and 4