Find the sum of the numbers lying between 1 and 100 which are divisible by 2 or 5.
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
Sir this question answer n=500,sn=27,54,000
To find the sum of the numbers between 1 and 100 that are divisible by 2 or 5, we can use the concept of arithmetic series.
First, let's find the numbers divisible by 2 between 1 and 100. We have two options to approach this problem:
Option 1:
We can generate a list of numbers between 1 and 100 and check if each number is divisible by 2. If it is, we add it to a running sum.
Option 2:
Since we know that every other number starting from 2 is divisible by 2, we can generate a list of numbers by starting from 2 and adding 2 repeatedly until we reach 100. Again, we add each number to a running sum.
For this particular problem, option 2 is more efficient. Using option 1 would still get us the correct answer, but it would require unnecessary computations.
Now, let's calculate the sum of these divisible numbers.
We can use the formula for the sum of an arithmetic series:
Sum = (n/2) * (first term + last term)
To find the number of terms (n), we can divide the difference between the first and last term by the common difference (in this case, 2). We subtract 2 from 100 (the upper limit) to ensure we don't count any numbers outside the range.
Using option 2 to generate the numbers divisible by 2, the first term would be 2, and the last term would be 100 (or the nearest multiple of 2 below 100). Plug these values into the formula:
n = (100 - 2) / 2 = 98 / 2 = 49
Now, we can calculate the sum of these divisible numbers:
Sum = (49/2) * (2 + 100) = 24.5 * 102 = 2499
Therefore, the sum of the numbers between 1 and 100 that are divisible by 2 is 2499.
Now, let's repeat the same process to find the sum of numbers divisible by 5.
Using option 2, the first term would be 5, and the last term would be 100 (or the nearest multiple of 5 below 100). Plug these values into the formula:
n = (100 - 5) / 5 = 95 / 5 = 19
Calculate the sum:
Sum = (19/2) * (5 + 100) = 9.5 * 105 = 997.5
Now, we need to account for the numbers that are divisible by both 2 and 5, which means we counted them twice.
Using the same method, we find that there are (100 - 10) / 10 = 9 terms that are divisible by both 2 and 5. The first term is 10, and the last term is 90.
Sum = (9/2) * (10 + 90) = 4.5 * 100 = 450
To get the final sum, we add the sum of the divisible numbers by 2 (2499) and the sum of the divisible numbers by 5 (997.5), and subtract the sum of the divisible numbers by both 2 and 5 (450):
Final Sum = 2499 + 997.5 - 450 = 3046.5
Therefore, the sum of the numbers lying between 1 and 100 which are divisible by 2 or 5 is 3046.5.
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