The sum of the first hundred even natural numbers divisible by 5 isa)5...
first even no. divisible by 5= 10
d=10
Sn = n/2(2a + (n-1) d)
S100= 100/2 (2(10) + (100-1)10)
S100=50(20 + 990)
S100= 50(1010)
S100 = 50500
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The sum of the first hundred even natural numbers divisible by 5 isa)5...
Calculation of sum of first hundred even natural numbers divisible by 5
Approach:
To find the sum of the first hundred even natural numbers divisible by 5, we need to first identify these numbers and then add them up.
Identification of even natural numbers divisible by 5:
Even natural numbers are those which are divisible by 2. Since we need even numbers that are also divisible by 5, we can multiply 5 with every even natural number starting from 1 till we get 100 such numbers.
Calculation of sum:
Once we have identified the hundred even natural numbers divisible by 5, we can add them up to get the sum.
Formula:
The sum of the first n even natural numbers is given by the formula n(n+1). Since we need the sum of the first 100 such numbers, we can use this formula to calculate the sum.
Calculation:
1. Identify the first hundred even natural numbers that are divisible by 5:
- 5 x 2 = 10
- 5 x 4 = 20
- 5 x 6 = 30
- ...
- 5 x 200 = 1000
2. Add up these hundred numbers:
- 10 + 20 + 30 + ... + 1000
3. Use the formula n(n+1) to calculate the sum:
- Sum = 100(100+1) = 10100
4. Since each of these hundred even natural numbers is multiplied by 5, we need to multiply the sum by 5 to get the final answer:
- Final Answer = 5 x 10100 = 50500
Answer:
Therefore, the sum of the first hundred even natural numbers divisible by 5 is 50500. Hence, option D is the correct answer.
The sum of the first hundred even natural numbers divisible by 5 isa)5...
This is an A.P..n we hve ..a,d as well as n n l..so we can also use d formula..Sn..=n/2(a+l)...if l is given then we shd use this formula...it will reduce our time naa...
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