Find the sum of integers from $1$ to $100$ that are divisible by $2$ or $5.$

Vedclass pdf generator app on play store
Vedclass iOS app on app store

The integers from $1$ to $100,$ which are divisible by $2,$ are $2,4,6 \ldots \ldots 100$

This forms an $A.P.$ with both the first term and common difference equal to $2.$

$\Rightarrow 100=2+(n-1) 2$

$\Rightarrow n=50$

$\therefore 2+4+6+\ldots \ldots+100=\frac{50}{2}[2(2)+(50-1)(2)]$

$=\frac{50}{2}[4+98]$

$=(25)(102)$

$=2550$

The integers from $1$ to $100 ,$ which are divisible by $5,10 \ldots . .100$

This forms an $A.P.$ with both the first term and common difference equal to $5 .$

$\therefore 100=5+(n-1) 5$

$\Rightarrow 5 n=100$

$\Rightarrow n=20$

$\therefore 5+10+\ldots .+100=\frac{20}{2}[2(5)+(20-1) 5]$

$=10[10+(19) 5]$

$=10[10+95]=10 \times 105$

$=1050$

The integers, which are divisible by both $2$ and $5,$ are $10,20, \ldots \ldots 100$

This also forms an $A.P.$ with both the first term and common difference equal to $10.$

$\therefore 100=10+(n-1)(10)$

$\Rightarrow 100=10 n$

$\Rightarrow n=10$

$\therefore 10+20+\ldots .+100=\frac{10}{2}[2(10)+(10-1)(10)]$

$=5[20+90]=5(110)=550$

$\therefore$ Required sum $=2550+1050-550=3050$

Thus, the sum of the integers from $1$ to $100,$ which are divisible by $2$ or $5,$ is $3050$

Similar Questions

A man deposited $Rs$ $10000$ in a bank at the rate of $5 \%$ simple interest annually. Find the amount in $15^{\text {th }}$ year since he deposited the amount and also calculate the total amount after $20$ years.

Let $a_1, a_2, a_3, \ldots, a_{100}$ be an arithmetic progression with $a_1=3$ and $S_p=\sum_{i=1}^p a_i, 1 \leq p \leq 100$. For any integer $n$ with $1 \leq n \leq 20$, let $m=5 n$. If $\frac{S_{m m}}{S_n}$ does not depend on $n$, then $a_2$ is

  • [IIT 2011]

Write the first five terms of the sequences whose $n^{t h}$ term is $a_{n}=(-1)^{n-1} 5^{n+1}$

If the $9^{th}$ term of an $A.P.$ be zero, then the ratio of its $29^{th}$ and $19^{th}$ term is

The ratio of the sums of $m$ and $n$ terms of an $A.P.$ is $m^{2}: n^{2} .$ Show that the ratio of $m^{ th }$ and $n^{ th }$ term is $(2 m-1):(2 n-1)$