Ap is 2,4,6....98
Tn= a + (n-1)d
98= 2 + 2n-2
n=48
Sn = n/2(a+Tn)
= 24(100)
=2400
Find the sum of the numbers lying between 1 and 100 which are divisible by 2 or 5.
3 answers
First of all we need all the even number, which are
2,4,6,..., 98
there are 49 of them, so
sum(49 evens) = (49/2)(4 + 2(48)) = 2450
Now this includes all multiples of 5 that end it 0, but not those that end in 5
so we have to add
5,15,25,35,45,55,65,75,85, and 95
they add up to 500 (I just added them)
The sum of your numbers = 2450+500 = 2950
2,4,6,..., 98
there are 49 of them, so
sum(49 evens) = (49/2)(4 + 2(48)) = 2450
Now this includes all multiples of 5 that end it 0, but not those that end in 5
so we have to add
5,15,25,35,45,55,65,75,85, and 95
they add up to 500 (I just added them)
The sum of your numbers = 2450+500 = 2950
Sir this question answer n=500,sn=27,54,000