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

Given, S₆ = 42 and a₁₀ : a₃₀ = 1 : 3

$\Rightarrow a${10} : a{30} = 1 : 3 \\[1em] \Rightarrow \dfrac{a{10}}{a{30}} = \dfrac{1}{3} \\[1em] \Rightarrow 3a{10} = a{30}

By formula, a_n = a + (n - 1)d,

$\Rightarrow 3[a + (10 - 1)d] = a + (30 - 1)d \\[1em] \Rightarrow 3(a + 9d) = a + 29d \\[1em] \Rightarrow 3a + 27d = a + 29d \\[1em] \Rightarrow 3a - a = 29d - 27d \\[1em] \Rightarrow 2a = 2d \\[1em] \Rightarrow a = d.$

S₆ = 42 or,

⇒ $\dfrac{6}{2}[2a + (6 - 1)d]$ = 42

Since, a = d
⇒ 3[2d + 5d] = 42
⇒ 3 × 7d = 42
⇒ 21d = 42
⇒ d = 2.

Since, a = d hence a = 2.

By formula, a_n = a + (n - 1)d
⇒ a₁₃ = 2 + (13 - 1)(2)
⇒ a₁₃ = 2 + 24
⇒ a₁₃ = 26.

Hence, the first term of the A.P. is 2 and the thirteenth term is 26.