Question

find three consecutive even integers such that the sum of the first and the last numbers exceeds the second no. by 10

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}3\mathrm{consecutive}\mathrm{even}\mathrm{integers}\mathrm{be}2\mathrm{x},2\mathrm{x}+2\mathrm{and}2\mathrm{x}+4. \\ \mathrm{Now},\mathrm{according}\mathrm{to}\mathrm{question}, \\ 2\mathrm{x}+\left(2\mathrm{x}+4\right)=\left(2\mathrm{x}+2\right)+10 \\ \Rightarrow 4\mathrm{x}+4=2\mathrm{x}+12 \\ \Rightarrow 4\mathrm{x}-2\mathrm{x}=12-4 \\ \Rightarrow 2\mathrm{x}=8 \\ \Rightarrow \mathrm{x}=4 \\ 2\mathrm{x}=2\times 4=8 \\ 2\mathrm{x}+2=8+2=10 \\ 2\mathrm{x}+4=8+4=12 \\ \mathrm{So},\mathrm{consecutive}\mathrm{integers}\mathrm{are}: \\ 8,10\mathrm{and}12 \end{gathered}$$