$\begin{array}{1 1}3050 \\ 3600 \\ 4150 \\ 3750 \end{array} $

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- $n^{th}$ term of an A.P.$t_n=a+(n-1)d$
- Sum of $n$ terms of the A.P. =$\large\frac{n}{2}$$(t_n+a)$

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

$=2550+1050-550=3050$.

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