Question

The sum of the first three numbers in an arithmetic progression is 18. If the product of the first and third term is 5 times the common difference, find the three numbers.

Answer 1

Let the first three numbers in an arithmetic progression be a − d, a, a + d.

The sum of the first three numbers in an arithmetic progression is 18.

a − d + a + a + d = 18
⇒ 3a = 18
⇒ a = 6

The product of the first and third term is 5 times the common difference.

$$\begin{gathered} \left(a-d\right)\left(a+d\right)=5d \\ \Rightarrow a^2-d^2=5d \\ \Rightarrow {\left(6\right)}^2-d^2=5d \\ \Rightarrow 36-d^2=5d \\ \Rightarrow d^2+5d-36=0 \\ \Rightarrow d^2+9d-4d-36=0 \\ \Rightarrow d\left(d+9\right)-4\left(d+9\right)=0 \\ \Rightarrow \left(d+9\right)\left(d-4\right)=0 \\ \Rightarrow d=4,-9 \\ \mathrm{For}d=4, \\ \mathrm{The}\mathrm{numbers}\mathrm{are}6-4,6,6+4 \\ \mathrm{i}.\mathrm{e}.2,6,10 \\ \mathrm{For}d=-9, \\ \mathrm{The}\mathrm{numbers}\mathrm{are}6+9,6,6-9 \\ \mathrm{i}.\mathrm{e}.15,6,-3 \end{gathered}$$

Hence, the three numbers are 2, 6, 10 or 15, 6, −3.