Find five numbers in A.P. whose sum is $12\dfrac{1}{2}$ and the ratio of the first to the last term is 2 : 3.

Let the numbers be (a - 2d), (a - d), a, (a + d), (a + 2d).

Given, sum = $12\dfrac{1}{2}$.

$\therefore a - 2d + a - d + a + a + d + a + 2d = 12\dfrac{1}{2} \\[1em] \Rightarrow 5a = \dfrac{25}{2} \\[1em] \Rightarrow a = \dfrac{5}{2} = 2.5.$

Given, ratio of the first to the last term is 2 : 3.

$\therefore \dfrac{a - 2d}{a + 2d} = \dfrac{2}{3} \\[1em] \Rightarrow 3(a - 2d) = 2(a + 2d) \\[1em] \Rightarrow 3a - 6d = 2a + 4d \\[1em] \Rightarrow 3a - 2a = 4d + 6d \\[1em] \Rightarrow a = 10d \\[1em] \Rightarrow \dfrac{5}{2} = 10d \\[1em] \Rightarrow d = \dfrac{5}{2 \times 10} \\[1em] \Rightarrow d = \dfrac{1}{4} = 0.25.$

Numbers = 2.5 - 2 × 0.25, 2.5 - 0.25, 2.5, 2.5 + 0.25, 2.5 + 2 × 0.25

= 2, 2.25, 2.5, 2.75, 3.

Hence, the numbers are 2, 2.25, 2.5, 2.75, 3.