- Trending Categories
- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

# Find the sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.

Given:

To do:

We have to find the sum of all

Solution:

Numbers that are multiples of 2 as well as 5 are the multiples of LCM of 2 and 5.

LCM of 2 and 5 $=2\times5=10$

Numbers divisible by 10 are $10, 20,....., 100, 110,....., 990, 1000,......$

Numbers divisible by 2 and 5 from 1 to 500 are $10, 20, ......,500$

This series is in A.P.

Here,

First term $a=10$

Common difference $d=10$

Last term $a_n=500$

We know that,

$a_n=a+(n-1)d$

$500=10+(n-1)10$

$500-10=(n-1)10$

$490=(n-1)10$

$49=n-1$

$n=49+1$

$n=50$

We know that,

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

$=\frac{50}{2}[2 \times 10+(50-1) \times 10]$

$=25[20+49 \times 10]$

$=25(20+490)$

$=25 \times 510$

$=12750$

The sum of all integers from 1 to 500 which are multiples of 2 as well as 5 is $12750$.

- Related Questions & Answers
- Sum of series 1^2 + 3^2 + 5^2 + . . . + (2*n – 1)^2
- Sum of series 1^2 + 3^2 + 5^2 + . . . + (2*n - 1)^2 in C++
- C++ program to find the sum of the series (1*1) + (2*2) + (3*3) + (4*4) + (5*5) + … + (n*n)
- Find sum of Series with n-th term as n^2 - (n-1)^2 in C++
- Sum of the Series 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + ... in C++
- C/C++ Program to Find sum of Series with n-th term as n power of 2 - (n-1) power of 2
- Java Program to Find sum of Series with n-th term as n^2 – (n-1)^2
- Find sum of the series 1-2+3-4+5-6+7....in C++
- Python Program for Find sum of Series with the n-th term as n^2 – (n-1)^2
- C/C++ Program to Find the sum of Series with the n-th term as n^2 – (n-1)^2
- C++ program to find the sum of the series 1 + 1/2^2 + 1/3^3 + …..+ 1/n^n
- Integers have sum of squared divisors as perfect square in JavaScript
- Sum of the series 1 + (1+3) + (1+3+5) + (1+3+5+7) + + (1+3+5+7+....+(2n-1)) in C++
- Sum of the series 1 + (1+3) + (1+3+5) + (1+3+5+7) + ...... + (1+3+5+7+...+(2n-1)) in C++
- Find Multiples of 2 or 3 or 5 less than or equal to N in C++
- Sum of the series 1 / 1 + (1 + 2) / (1 * 2) + (1 + 2 + 3) / (1 * 2 * 3) + … + upto n terms in C++