Question

Michael wrote all the two digit numbers from 10 to 99. He calculated the sum of all those numbers which were divisible by 5. The sum obtained was equal to

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{list}\mathrm{of}2-\mathrm{digit}\mathrm{numbers}\mathrm{from}10\mathrm{to}99,\mathrm{that}\mathrm{are}\mathrm{divisible}\mathrm{by}5\mathrm{is}: \\ 10,15,20,25,............95. \\ \mathrm{The}\mathrm{numbers}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{sequence}\mathrm{is}\mathrm{in}\mathrm{AP},\mathrm{with} \\ \mathrm{first}\mathrm{term},\mathrm{a}=10\mathrm{and}\mathrm{common}\mathrm{difference},\mathrm{d}=5 \\ \mathrm{Let}\mathrm{n}\mathrm{be}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{terms}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{AP}. \\ \mathrm{now},{\mathrm{a}}_{\mathrm{n}}=95 \\ \Rightarrow \mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}=95 \\ \Rightarrow 10+\left(\mathrm{n}-1\right)5=95 \\ \Rightarrow 5\mathrm{n}-5=85 \\ \Rightarrow 5\mathrm{n}=90 \\ \Rightarrow \mathrm{n}=18 \\ \mathrm{Now}, \\ \mathrm{Sum}\mathrm{of}\mathrm{first}\mathrm{n}\mathrm{terms}\mathrm{of}\mathrm{an}\mathrm{AP}\mathrm{is}\mathrm{given}\mathrm{by}, \\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right] \\ {\mathrm{S}}_{18}=\frac{18}{2}\left[2\times 10+17\times 5\right]=9\left(20+85\right)=9\times 105=945 \end{gathered}$$