The sum of first six terms of an AP is 42. The ratio of its 10th term to its 30th term is 1:3. Calculate the 1st and thirteenth term of the AP.

Let a₁  = the first term....we have....the sum of the 1st 6 terms  = 42....therefore...

42 = (6/2)[2a₁ + 5d]

42 = 3[2a₁ + 5d]

14 = 2a₁ + 5d  →   a₁ = [14 - 5d]/2 

And a₃₀ = 3(a₁₀)   .... therefore.....

3(a₁₀) = a₁₀ + 20d

2(a₁₀) = 20d

a₁₀ = 10d

So we have

a₁₀ = (a₁ + 9d)  ... and by substitution.....

10d = ([14- 5d]/2 + 9d)

20d = 14 - 5d + 18d

20d = 14 + 13d

7d = 14

d = 2

Therefore....

a₁ = [14 - 5(2)] / 2  = [14 - 10] / 2 =  4/2 = 2

And

a₁₃ = [2 + 2(12)] =  26