Question

Find five numbers in an AP whose sum is 25 and product is 135

Answer 1

$$\begin{gathered} \mathrm{Let}5\mathrm{numbers}\mathrm{be}(\mathrm{a}-2\mathrm{d}),(\mathrm{a}-\mathrm{d}),\mathrm{a},(\mathrm{a}+\mathrm{d}),(\mathrm{a}+2\mathrm{d}) \\ \mathrm{Sum}\mathrm{of}\mathrm{numbers}=25 \\ \Rightarrow \mathrm{a}-2\mathrm{d}+\mathrm{a}-\mathrm{d}+\mathrm{a}+\mathrm{a}+\mathrm{d}+\mathrm{a}+2\mathrm{d}=25 \\ \Rightarrow 5\mathrm{a}=25 \\ \Rightarrow \mathrm{a}=5 \\ \mathrm{Product}=135 \\ \Rightarrow (\mathrm{a}-2\mathrm{d})(\mathrm{a}-\mathrm{d})\mathrm{a}(\mathrm{a}+\mathrm{d})(\mathrm{a}+2\mathrm{d})=135 \\ \Rightarrow (5-2\mathrm{d})(5-\mathrm{d})5(5+\mathrm{d})(5+2\mathrm{d})=135 \\ \Rightarrow (5^2-(2\mathrm{d}{)}^2\left)\right(5^2-{\mathrm{d}}^2)=27 \\ \Rightarrow (25-4{\mathrm{d}}^2\left)\right(25-{\mathrm{d}}^2)=27 \\ \Rightarrow 625-25{\mathrm{d}}^2-100{\mathrm{d}}^2+4{\mathrm{d}}^4=27 \\ \Rightarrow 4{\mathrm{d}}^4-125{\mathrm{d}}^2+598=0 \\ \mathrm{Let}{\mathrm{d}}^2=\mathrm{t} \\ 4{\mathrm{t}}^2-125\mathrm{t}+598=0 \\ \mathrm{Solving}\mathrm{Q}.\mathrm{E}\mathrm{in}\mathrm{t},\mathrm{we}\mathrm{have} \\ \mathrm{t}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}} \\ \mathrm{t}=\frac{125\pm \sqrt{{125}^2-4\left(4\right)\left(598\right)}}{2\times 4} \\ \mathrm{t}=25.35,5.89 \\ \therefore \mathrm{d}=\sqrt{\mathrm{t}} \\ \mathrm{d}=\sqrt{25.35}\mathrm{or}\mathrm{d}=\sqrt{5.89} \\ \mathrm{Numbers}\mathrm{are}:(5-2\sqrt{25.35}),(5-\sqrt{25.35}),5,(5+\sqrt{25.35}),(5+2\sqrt{25.35}) \\ \mathrm{or}(5-2\sqrt{5.89}),(5-\sqrt{5.89}),5,(5+\sqrt{5.89}),(5+2\sqrt{5.89}) \end{gathered}$$