Numbers from $1$ and $100$ that are divisible by $2$ and $5$ are
$2,4,5,6,8,10,12,14,15,...........$
These numbers are grouped into three groups,
$2,4,6,8,10,...........100$...........(i)
$5,10,15,20,........100$........(ii) and
$10,20,30.........100$.......(iii)
Here group (iii) contains numbers that are common in both group (i) and (ii)
$\therefore\:Sum of the required numbers =
$um of (i) + sum of (ii) - sum of (iii).
(i) is an A.P. with $1^{st}$ term $a=2$, common difference $d=2$ and $t_n=100$
$\Rightarrow\:a+(n-1)d=100$
$\Rightarrow\:2+(n-1)2=100$ or $n-1=49$ or $n=50$
$\therefore$ Sum of the numbers in ((i) = $\large\frac{n}{2}$$(t_n+a)=25(100+2)=2550$.
Step 2
(ii) is an A.P. with $a=5$, $d=5$ and $t_n=100$
$\Rightarrow\:a+(n-1)d=100$
$\Rightarrow\:5+(n-1)5=100$ or $ n=19+1=20$
$\therefore$ Sum of the numbers in (ii) $=\large\frac{n}{2}$$(t_n+a)=10(100+5)=1050.$
Step 3
Similarly (ii) is also an A.P. with $a=10$, $d=10$ and $t_n=100$
$\Rightarrow\:a+(n-1)d=100$
$\Rightarrow\:10+(n-1)10=100$ or n=10$
$\therefore$ Sum of the numbers in (iii) $=\large\frac{n}{2}$$(t_n+a)=5(100+10)=550$
Step 4
$\therefore\:$ Sum of the required numbers
$=2550+1050-550=3050$.