find the sum of integers between 200and 300 divisible by 5
Answers
Step-by-step explanation:
Given :-
Integers between 200 and 300
To find:-
Find the sum of integers between 200and 300 divisible by 5 ?
Solution :-
The list of numbers between 200 and 300
= 201, 202 , 203 , ..., 299
The list of numbers between 200 and 300 which are divisible by 5
= 205, 210 , 215 , ..., 295
First term = (a) = 205
Common difference =
d = 210-205 = 5
d = 215-210 = 5
Since the common difference is same throughout the series then they are in the AP.
Now we have to find
The sum of integers between 200and 300 divisible by 5
= 205 + 210 + 215 +...+ 295
Last term =(l)= 295
Let nth term = 295
We know that
nth term of the AP = an = a+(n-1)d
=> 205+(n-1)(5)=295
=> 205+5n-5 = 295
=> 200+5n = 295
=> 5n = 295-200
=> 5n = 95
=> n = 95/5
=> n = 19
Number of terms = 19
We know that
The sum of first n terms in an AP
= Sn = (n/2)[a+l]
We have
a = 205, l = 295 and n = 19
=> S19 = (19/2)[205+295]
=> S 19 = (19/2)(500)
=> S 19 = 19×250
=> S 19 = 4750
or
The sum of first n terms in an AP
= Sn = (n/2)[2a+(n-1)d]
=> S 19 = (19/2)[2(205)+(19-1)(5)]
=> S 19 = (19/2)[410+18(5)]
=> S 19 = (19/2)(410+90)
=> S 19 = (19/2)(500)
=> S 19 = 19×250
=> S 19 = 4750
Answer:-
The sum of the integers between 200 and 2
300 which are divisible by 5 is 4750
Used formulae:-
- nth term of the AP = an = a+(n-1)d
- The sum of first n terms in an AP
- = Sn = (n/2)[2a+(n-1)d]
- The sum of first n terms in an AP
- = Sn = (n/2)[a+l]