Question

Find three numbers in AP whose sum is 15 and whose product is 105.

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 15.

a − d + a + a + d = 15
⇒ 3a = 15
⇒ a = 5

Their product is 105.

$$\begin{gathered} \left(a-d\right)\times a\times \left(a+d\right)=105 \\ \Rightarrow \left(a^2-d^2\right)\times a=105 \\ \Rightarrow \left({\left(5\right)}^2-d^2\right)\times 5=105 \\ \Rightarrow {\left(5\right)}^2-d^2=21 \\ \Rightarrow 25-d^2=21 \\ \Rightarrow d^2=25-21 \\ \Rightarrow d^2=4 \\ \Rightarrow d=2,-2 \\ \mathrm{For}d=2, \\ \mathrm{The}\mathrm{numbers}\mathrm{are}5-2,5,5+2 \\ \mathrm{i}.\mathrm{e}.3,5,7 \\ \mathrm{For}d=-2, \\ \mathrm{The}\mathrm{numbers}\mathrm{are}5+2,5,5-2 \\ \mathrm{i}.\mathrm{e}.7,5,3 \end{gathered}$$

Hence, the three numbers are 3, 5, 7 or 7, 5, 3.