Question

Find five number in A.P such that their sumis 30 and the sum of their squares is 220?

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{a}-3\mathrm{d},\mathrm{a}-\mathrm{d},\mathrm{a},\mathrm{a}+\mathrm{d}\mathrm{and}\mathrm{a}+3\mathrm{d}\mathrm{be}\mathrm{the}\mathrm{five}\mathrm{numbers}\mathrm{in}\mathrm{AP}. \\ \mathrm{Now},\mathrm{sum}\mathrm{of}5\mathrm{numbers}=30 \\ \Rightarrow \mathrm{a}-3\mathrm{d}+\mathrm{a}-\mathrm{d}+\mathrm{a}+\mathrm{a}+\mathrm{d}+\mathrm{a}+3\mathrm{d}=30 \\ \Rightarrow 5\mathrm{a}=30 \\ \Rightarrow \mathrm{a}=6 \\ \mathrm{Now},{\left(\mathrm{a}-3\mathrm{d}\right)}^2+{\left(\mathrm{a}-\mathrm{d}\right)}^2+{\mathrm{a}}^2+{\left(\mathrm{a}+\mathrm{d}\right)}^2+{\left(\mathrm{a}+3\mathrm{d}\right)}^2=220 \\ \Rightarrow {\left(6-3\mathrm{d}\right)}^2+{\left(6-\mathrm{d}\right)}^2+6^2+{\left(6+\mathrm{d}\right)}^2+{\left(6+3\mathrm{d}\right)}^2=220 \\ \Rightarrow 36+9{\mathrm{d}}^2-36\mathrm{d}+36+{\mathrm{d}}^2-12\mathrm{d}+36+36+{\mathrm{d}}^2+12\mathrm{d}+36+9{\mathrm{d}}^2+36\mathrm{d}=220 \\ \Rightarrow 180+20{\mathrm{d}}^2=220 \\ \Rightarrow 20{\mathrm{d}}^2=40 \\ \Rightarrow {\mathrm{d}}^2=\frac{40}{20}=2 \\ \Rightarrow \mathrm{d}=\pm \sqrt{2} \\ \mathrm{When}\mathrm{d}=\sqrt{2},\mathrm{then}\mathrm{the}\mathrm{numbers}\mathrm{are}: \\ 6-3\sqrt{2},6-\sqrt{2},6,6+\sqrt{2},6+3\sqrt{2} \\ \mathrm{When}\mathrm{d}=-\sqrt{2},\mathrm{then}\mathrm{the}\mathrm{numbers}\mathrm{are}: \\ 6+3\sqrt{2},6+\sqrt{2},6,6-\sqrt{2},6-3\sqrt{2} \end{gathered}$$