The 19th term of an A.P. is equal to three times its 6th term. If its 9th term is 19, find the A.P.

We know that

    a_n = a + (n - 1)d

According to question,

⇒ a₁₉ = 3(a₆)
⇒ a + 18d = 3(a + 5d)
⇒ a + 18d = 3a + 15d
⇒ 3a - a = 18d - 15d
⇒ 2a = 3d
⇒ a = $\dfrac{3}{2}d$     (Eq 1)

Given, a₉ = 19

$\Rightarrow a + 8d = 19 \\[1em] \Rightarrow \dfrac{3}{2}d + 8d = 19 \\[1em] \Rightarrow \dfrac{3d + 16d}{2} = 19 \\[1em] \Rightarrow \dfrac{19d}{2} = 19 \\[1em] \Rightarrow d = \dfrac{19}{19} \times 2 \\[1em] \Rightarrow d = 2. \\[1em]$

Putting value of d in Eq 1 we get,

$\Rightarrow a = \dfrac{3}{2}d \\[1em] \Rightarrow a = \dfrac{3}{2} \times 2 \\[1em] \Rightarrow a = 3.$

Hence, the terms of A.P. are

a₂ = a₁ + d = 3 + 2 = 5,
a₃ = a₂ + d = 5 + 2 = 7,
a₄ = a₃ + d = 7 + 2 = 9.

Hence, the terms of the A.P. are 3, 5, 7, 9, ….