Question

Find the sum of all odd 3 digit numbers which are divisible by 5.

Answer 1

The sum of all odd three digit numbers which are divisible by 5 is 49500.

Step-by-step explanation:

The three digit numbers which are divisible by 5 are

105, 115, 125, 135, ... ..., 995

This forms an Arithmetic Progression, whose

• first term = 105
• common difference = 115 - 105 = 10
• last term = 995

Let, 995 be the nth term of the progression.

So, nth term = first term + (n - 1) × common difference

⇒ 995 = 105 + (n - 1) × 10

⇒ 995 = 105 + 10n - 10

⇒ 995 = 95 + 10n

⇒ 10n = 900

⇒ n = 90

Thus there are 90 terms in our required progression.

Now, the required sum

= $\dfrac{n}{2}$ (first term + last term)

= $\dfrac{90}{2}$ (105 + 995)

= 45 × 1100

= 49500

#SPJ3

Answer 2

Answer:

Step-by-step explanation:

Concept Introduction:

Here we will use Arithmetic Progression Concept to solve the question.

Given Information:

The numbers whose sum will be divisible by $5$ are all odd $3$ digit numbers.

To Find:

We have to find the sum of all odd $3$ digit numbers which are divisible by $5$.

Solution:

We need all 3 digit numbers i.e. least range is $100$ and best range is $999$.

Numbers should be divisible by $5$ .So,

$a_{1}=100 \\\\ a_{n}=995 \\ d=5$

The nth term of A.P. is,

$a_{n}=a_{1}+(n-1) d$

By substituting the value in this equation, we will get

$995=100+(n-1) 5$

The value of n is

$5(n-1)=995-100 5(n-1)=895 \\ n-1=179 \\ n=180$

The Sum of n terms of A.P. is,

$\sum a=\frac{n}{2}[2 a+(n-1) d]$

Substitute the values in the equation to get the value .

$\therefore \sum \mathrm{a}=\frac{180}{2}[(2 \times 100)+(180-1) 5] \\ \therefore \sum \mathrm{a}=90[200+(179 \times 5)] \\ \therefore \sum \mathrm{a}=90[200+895] \\ \therefore \sum \mathrm{a}=90 \times 1095 \\ \therefore \sum \mathrm{a}=98550$

The sum of all odd $3$ digit numbers is $98550$ which are divisible by $5$ .

#SPJ2