find the sum of all three digit natural numbers which are in the form 3m+2
Answers
Step-by-step explanation:
Given :-
three digit natural numbers which are in the form 3m+2
To find:-
Find the sum of all three digit natural numbers which are in the form 3m+2 ?
Solution:-
Given that
Three digit natural numbers which are in the form of 3m+2
=> three digit numbers leaving remainder 2 which are divided by 3
now
The list of three digit numbers = 100 ,101, ...., 999
The list of three digits numbers which are divisible by 3 = 102, 105, 108,..., 996,999
The list of three digits numbers which are divisible by 3 and leaves the remainder 2 .i.e. 3m+2 form numbers
= 104,107,110,113.., 998
First term = 104
Common difference = 107-104 =3
=> 110-107 = 3
Since the common difference is same throughout the series
So , The list of the numbers are in the AP
The general term of an AP = an = a+(n-1)d
We have
a = 104
d = 3
an = 998
=> 104+(n-1)(3)=998
=> 104+3n-3 = 998
=> 101+3n = 998
=> 3n = 998-101
=> 3n = 897
=> n = 897/3
=> n = 299
Number of terms = 299
So we have n = 299
We know that
Sum of first n natural numbers in an AP
S n = (n/2)[2a+(n-1)d]
=> S 299 = (299/2)[2(104)+(299-1)(3)]
=> S 299 = (299/2)[208+(298)(3)]
=> S 299 = (299/2)[208+894]
=> S 299 = (299/2)×(1102)
=> S 299 = 299×551
=> S 299 = 164749
Answer:-
The sum of all three digit natural numbers which are in the form 3m+2 is 164749
Used formulae:-
The general term of an AP = an = a+(n-1)d
Sum of first n natural numbers in an AP
S n = (n/2)[2a+(n-1)d]