Sum of all natural numbers between 40 and 400which are divisible by 7
Answers
Step-by-step explanation:
a=42
an=399
n=?
d=7
an=a+(n-1)×d
399=42+(n-1)×7
357=7n-7
n=50
Sn=n/2(a+an)
=50/2(42+399)
=25(42+399)
=11025
Answer:
The sum of natural number between 40 and 400 which are divisible by 7 is 11466.
Step-by-step explanation:
I believe your Question was,
"Find the Sum of all the natural numbers between 40 and 400 which are divisible by 7."
Here it forms an Arithmetic Progression(AP),
We know that,
The 1st number after 40 which is divisible by 7 is 42 then 49 and the last number which is divisible by 7 before 400 is 399.
Hence,
a(1st term) = 42
d(common difference) = 49 - 42 = 7
Remember, the numbers are divisible by 7 so they must have a difference of 7
l (last term) = 399
We know that,
a(nth) = a + (n - 1)d
a(nth) = l = 399
399 = 42 + (n - 1)7
399 - 42 = (n - 1)7
n - 1 = 357/7
n = 51 + 1
n = 52
Thus, there are 52 natural numbers between 40 and 400 which are divisible by 7
Now,
Sn = (n/2)[a + l]
S(52) = (52/2)[42 + 399]
S(52) = 26 × 441
S(52) = 11466
Thus, the sum of natural number between 40 and 400 which are divisible by 7 is 11466.
Hope it helped and you understood it........All the best