Question

Three numbers are in A.P. and their sum is 15. If 1, 3, 9 be added to them respectively, they form a G.P. Find the numbers.

Answer 1

Let the first term of an A.P. be a and its common difference be d.

$$\begin{gathered} a_1+a_2+a_3=15 \\ \Rightarrow a+\left(a+d\right)+\left(a+2d\right)=15 \\ \Rightarrow 3a+3d=15 \\ \Rightarrow a+d=5.......\left(\mathrm{i}\right) \\ \mathrm{Now},\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{question}: \\ \mathrm{a}+1,\mathrm{a}+\mathrm{d}+3\mathrm{and}\mathrm{a}+2\mathrm{d}+9\mathrm{are}\mathrm{in}\mathrm{G}.\mathrm{P}. \\ \Rightarrow {\left(\mathrm{a}+\mathrm{d}+3\right)}^2=\left(\mathrm{a}+1\right)\left(\mathrm{a}+2\mathrm{d}+9\right) \\ \Rightarrow {\left(5-\mathrm{d}+\mathrm{d}+3\right)}^2=\left(5-\mathrm{d}+1\right)\left(5-\mathrm{d}+2\mathrm{d}+9\right)\left[\mathrm{From}\left(\mathrm{i}\right)\right] \\ \Rightarrow {\left(8\right)}^2=\left(6-\mathrm{d}\right)\left(14+\mathrm{d}\right) \\ \Rightarrow 64=84+6d-14d-d^2 \\ \Rightarrow d^2+8d-20=0 \\ \Rightarrow \left(d-2\right)\left(d+10\right)=0 \\ \Rightarrow d=2,-10 \\ \mathrm{Now},\mathrm{putting}\mathrm{d}=2,-10\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get},a=3,15,\mathrm{respectively}. \\ \mathrm{Thus},\mathrm{for}a=3\mathrm{and}d=2,\mathrm{the}A.\mathrm{P}.\mathrm{is}3,5,7. \\ \mathrm{And},\mathrm{for}a=15\mathrm{and}d=-10,\mathrm{the}\mathrm{A}.\mathrm{P}.\mathrm{is}15,5,-5. \end{gathered}$$