find the sum of all natural numbers between 100 and 100 which are not divisible by 11
Answers
Answer:
44550 is the sum of all numbers between 100 and 1000 which are divisible by 11.
Given:
Numbers between 100 and 1000
To find:
Sum of all number between 100 and 1000 which are divisible by 11.
Solution:
The first number after 100 that is divisible by 11 is 111; therefore, the first number in the series is 111.
The last number or the number before 1000 that is a multiple of 11 is 999
Hence to find the sum of all the numbers that are divisible by 11 we are going to use the Sum of Arithmetic Progression method, but first let us find the number of such terms that are divisible by 11
\begin{gathered}\begin{array}{l}{T_{n}=a+(n-1) d} \\ {999=111+(n-1) 11} \\ {\frac{888}{11}=(n-1)} \\ {\frac{888}{11}+1=n, \quad n=81}\end{array}\end{gathered}
T
n
=a+(n−1)d
999=111+(n−1)11
11
888
=(n−1)
11
888
+1=n,n=81
Therefore, the sum of the numbers between 100 and 1000 is
\begin{gathered}\begin{array}{l}{S_{n}=\frac{n}{2}(2 a+(n-1) d)} \\ {S_{n}=\frac{81}{2}(550 \times 2)=44550}\end{array}\end{gathered}
S
n
=
2
n
(2a+(n−1)d)
S
n
=
2
81
(550×2)=44550
Answer:
1,2,3,4,5,6,7,8,9,10,12,13,14,15,16,17,18,19,20,21,23,24,25,26,27,28,29,30,31,32,34,35,36,37,38,29,40,41,42,43,45,46,47,48,49,50,51,52,53,54,56,57,58,59,60,61,62,63,64,65,67,68,69,70,71,72,73,74,75,76,78,79,80,81,82,83,84,85,86,87,89,90,91,92,93,94,95,96,97,98,100.
Step-by-step explanation: