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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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Find the sum of all the integers from $1$ and $100$ that are divisible by $2$ or $5$.

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

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1 Answer

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Toolbox:
  • $n^{th}$ 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
$=2550+1050-550=3050$.
answered Mar 29, 2014 by rvidyagovindarajan_1
edited Mar 29, 2014 by rvidyagovindarajan_1
 

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