Question

The sum of three numbers a, b, c in A.P. is 18. If a and b are each increased by 4 and c is increased by 36, the new numbers form a G.P. Find a, b, c.

Answer 1

Let the first term of the A.P. be a and the common difference be d.
∴ a = a , b = a + d and c = a + 2d
$$\begin{gathered} a+b+c=18 \\ \Rightarrow a+\left(a+d\right)+\left(a+2d\right)=18 \\ \Rightarrow 3a+3d=18 \\ \Rightarrow a+d=6.......\left(\mathrm{i}\right) \\ \mathrm{Now},\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{question},\mathrm{a}+4,\mathrm{a}+\mathrm{d}+4\mathrm{and}\mathrm{a}+2\mathrm{d}+36\mathrm{are}\mathrm{in}\mathrm{G}.\mathrm{P}. \\ \therefore {\left(\mathrm{a}+\mathrm{d}+4\right)}^2=\left(\mathrm{a}+4\right)\left(\mathrm{a}+2\mathrm{d}+36\right) \\ \Rightarrow {\left(6-\mathrm{d}+\mathrm{d}+4\right)}^2=\left(6-\mathrm{d}+4\right)\left(6-d+2\mathrm{d}+36\right) \\ \Rightarrow {\left(10\right)}^2=\left(10-\mathrm{d}\right)\left(42+\mathrm{d}\right) \\ \Rightarrow 100=420+10d-42d-d^2 \\ \Rightarrow d^2+32d-320=0 \\ \Rightarrow \left(d+40\right)\left(d-8\right)=0 \\ \Rightarrow d=8,-40 \\ \mathrm{Now},\mathrm{putting}d=8,-40\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get},a=-2,46,\mathrm{respectively}. \\ \mathrm{For}a=-2\mathrm{and}d=8,\mathrm{we}\mathrm{have}: \\ a=-2,b=6,c=14 \\ \mathrm{And},\mathrm{for}a=46\mathrm{and}d=-40,\mathrm{we}\mathrm{have}: \\ a\mathit{}=\mathit{}\mathit{46}\mathit{}\mathit{,}\mathit{}b\mathit{}=\mathit{}\mathit{6}\mathit{}\mathit{,}\mathit{}c\mathit{}=\mathit{}\mathit{-}\mathit{34} \end{gathered}$$