The sum of integers from 1 to 100 that are di | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) To find the sum of integers from $1$ to $100$ divisible by $2$ or $5$,we use the Principle of Inclusion-Exclusion: $S = S_2 + S_5 - S_{10}$,where

Vedclass · Accepted Answer

$3050$

Vedclass · Answer

(B) To find the sum of integers from $1$ to $100$ divisible by $2$ or $5$,we use the Principle of Inclusion-Exclusion: $S = S_2 + S_5 - S_{10}$,where $S_n$ is the sum of integers divisible by $n$.$1$. Sum of integers divisible by $2$ $(2, 4, ..., 100)$: This is an arithmetic progression with $n = 50$,$a = 2$,$l = 100$. Sum $= \frac{50}{2}(2 + 100) = 25 	imes 102 = 2550$.$2$. Sum of integers divisible by $5$ $(5, 10, ..., 100)$: This is an arithmetic progression with $n = 20$,$a = 5$,$l = 100$. Sum $= \frac{20}{2}(5 + 100) = 10 	imes 105 = 1050$.$3$. Sum of integers divisible by both $2$ and $5$ (i.e.,divisible by $10$) $(10, 20, ..., 100)$: This is an arithmetic progression with $n = 10$,$a = 10$,$l = 100$. Sum $= \frac{10}{2}(10 + 100) = 5 	imes 110 = 550$.Total sum $= 2550 + 1050 - 550 = 3050$.

The sum of integers from $1$ to $100$ that are divisible by $2$ or $5$ is

Similar Questions

If ${a_1}, {a_2}, {a_3}, ..., {a_{24}}$ are in arithmetic progression and ${a_1} + {a_5} + {a_{10}} + {a_{15}} + {a_{20}} + {a_{24}} = 225$,then ${a_1} + {a_2} + {a_3} + ... + {a_{23}} + {a_{24}} = $

If $x, y, z$ are in $G.P.$ and $a^x = b^y = c^z$,then

The $A.M., G.M.$ and $H.M.$ between two numbers are $\frac{144}{15}$,$15$ and $12$,but not necessarily in this order. Then $H.M., G.M.$ and $A.M.$ respectively are

Find the $n^{th}$ term of the series:
$\frac{1}{n} + \frac{n+1}{n} + \frac{2n+1}{n} + \ldots$

The sum of all the digits of the numbers from $1$ to $100$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module