Question

In an arthmetic progression the sum of first 6 terms is 42. The ratio of 10th and 30th term is 1 :3 .find its 1st and 13th term.

Answer 1

1st term = 2
13th term = 26

here first term(a) and the common difference(d) will be equal i.e  equal to 2.
(to get this use the ratio given).

Answer 2

Sum of first 6 terms= 42
a10/a30= 1/3
3*a10=a30
3*(a+9d)=a+29d
3a+27d=a+29d
2a=2d
a=d

S=n/2[2a+(n-1)d]
42=6/2[2a+5d]
42=3(2a+5a)        since a=d
42=21a
a=2

d=2

a13=a+12d
      =2+12*2
       =26

hope its helpful
plzz mark brainiest