(c) Find the sum of all 2-digits natural numbers divisible by 3.
Answers
Step-by-step explanation:
Given:-
2-digits natural numbers
To find:-
Find the sum of all 2-digits natural numbers divisible by 3 ?
Solution:-
The list of all 2-digits natural numbers
= 10,11,12,13,...,99
The list of all 2-digit natural numbers which are divisible by 3
=> Multiples of 3 in 2- digit numbers
=> 12,15,18,...,99
First term of the list = 12
difference between two consecutive numbers
= 15-12=3
=18-15=3
=99-96 = 3
Common difference = 3
The Common difference is same throughout the series
So , 12,15,18,...99 are in the AP
We have
First term (a) = 12
Common difference (d) = 3
Last term (an)=99
We know that.
The general term or nth term or last term in an AP
is an = a+(n-1)d
=>12+(n-1)(3) = 99
=> 12+3n-3 = 99
=> 9+3n = 99
=> 3n = 99-9
=> 3n = 90
=> n = 90/3
=> n = 30
Number of terms in the AP = 30
Now ,
Sum of all terms in the AP
12+15+18+...+99
We know that
The sum of the first n terms in an AP is
Sn = (n/2)[2a+(n-1) d]
On Substituting these values in the above formula
=> S 30 = (30/2)[2(12)+(30-1)(3)]
=> S 30 = (15)[24+(29)(3)]
=> S 30= (15)[24+87]
=> S 30 = (15)(111)
=> S 30 = 1665
(or)
Sum of the first n terms = Sn = (n/2)(a+an)
=> S 30 = (30/2)(12+99)
=> S 30 = (15)(111)
=> S 30 = 1665
Answer:-
The sum of all 2-digits natural numbers divisible by 3 is 1665
Used formulae:-
- The general term or nth term or last term in an AP is an = a+(n-1)d
- Sum of the first n terms =
- Sn = (n/2)(a+an)
- Sn = (n/2)[2a+(n-1) d]
- a = first term
- d = Common difference
- n = Number of terms