The sum of all integers between 1 and 100 (bo | vedclass

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

Question

(A) Let $S_5$ be the sum of integers divisible by $5$ in the range $[1, 100]$. These are $5, 10, \dots, 100$. This is an arithmetic progression with $

Vedclass · Accepted Answer

$1349$

Vedclass · Answer

(A) Let $S_5$ be the sum of integers divisible by $5$ in the range $[1, 100]$. These are $5, 10, \dots, 100$. This is an arithmetic progression with $a = 5$,$l = 100$,and $n = \frac{100}{5} = 20$. The sum $S_5 = \frac{20}{2}(5 + 100) = 10 	imes 105 = 1050$.Let $S_{13}$ be the sum of integers divisible by $13$ in the range $[1, 100]$. These are $13, 26, 39, 52, 65, 78, 91$. Here $n = 7$. The sum $S_{13} = \frac{7}{2}(13 + 91) = \frac{7}{2}(104) = 7 	imes 52 = 364$.Let $S_{65}$ be the sum of integers divisible by both $5$ and $13$ (i.e.,divisible by $65$). The only such integer is $65$. So $S_{65} = 65$.By the Principle of Inclusion-Exclusion,the required sum is $S = S_5 + S_{13} - S_{65} = 1050 + 364 - 65 = 1349$.

The sum of all integers between $1$ and $100$ (both inclusive) which are divisible by $5$ or $13$ is

Similar Questions

If the sum of the first $n$ terms of a sequence is of the form $An^2 + Bn$,where $A$ and $B$ are constants independent of $n$,then the sequence is a ........

If $S_n = nP + \frac{1}{2}n(n - 1)Q$,where $S_n$ denotes the sum of the first $n$ terms of an $A.P.$,then the common difference is

The sum of the first four terms of an $A.P.$ is $56$. The sum of the last four terms is $112$. If its first term is $11$,the number of terms is

Write the first five terms of the sequence whose $n^{th}$ term is $a_{n} = \frac{2n - 3}{6}$.

The sum of the first $n$ natural numbers is

Vedclass Test Series

Exam Paper Generator

Online Exam Module