Three numbers are in A.P. and their sum is 15. If 1, 4 and 19 are added to these numbers respectively, the resulting numbers are in G.P. Find the numbers.

Let three numbers that are in A.P. be a - d, a, a + d.

Given, sum of three numbers = 15.

⇒ a - d + a + a + d = 15
⇒ 3a = 15
⇒ a = 5.

By adding 1, 4 and 19 in the terms become,

⇒ a - d + 1, a + 4 and a + d + 19
⇒ 5 - d + 1, 5 + 4 and 5 + d + 19
⇒ 6 - d, 9 and d + 24.

According to question these terms become in G.P.,

$\therefore \dfrac{9}{6 - d} = \dfrac{d + 24}{9} \\[1em] \Rightarrow (d + 24)(6 - d) = 81 \\[1em] \Rightarrow 6d - d^2 + 144 - 24d = 81 \\[1em] \Rightarrow 6d - d^2 - 24d + 144 - 81 = 0 \\[1em] \Rightarrow -18d - d^2 + 63 = 0 \\[1em] \Rightarrow d^2 + 18d - 63 = 0 \\[1em] \Rightarrow d^2 + 21d - 3d - 63 = 0 \\[1em] \Rightarrow d(d + 21) - 3(d + 21) = 0 \\[1em] \Rightarrow (d - 3)(d + 21) = 0 \\[0.5em] \Rightarrow d - 3 = 0 \text{ or } d + 21 = 0 \\[1em] \Rightarrow d = 3 \text{ or } d = -21.$

Taking d = 3,

∴ a - d = 5 - 3 = 2, a = 5, a + d = 5 + 3 = 8.

Taking d = 21,

∴ a - d = 5 - 21 = -16, a = 5, a + d = 5 + 21 = 26.

Hence, the required numbers are 2, 5 and 8 or -16, 5 and 26.