what is the sum of all natural numbers from 1 to 100 that are divisible by 7
Answers
Answer:
735
Step-by-step explanation:
It's a Arithmetic Progression
a = 7
d = 7
n = 14
Using AP Sum Formula = n*(2a + (n-1)d)/2 = 735
The sum of all the natural numbers from 1 to 100 that are divisible by 7 is 735.
Given: The range of 1 to 100.
To Find: The sum of all natural numbers from 1 to 100 that are divisible by 7
Solution:
- We can say that we need to find all the multiples of 7 in the range of 1 to 100.
- So, this is an Arithmetic progression starting from 7 and ending at the last multiple of 7 ( 98 ) within 100. So, we need to find the number of terms which can be done using the formula,
tn = a + (n – 1) × d ... (1)
where tn = last term, a = first term, n = number of terms, d = common difference.
- We need to find the sum of the A.P which can be done using the formula,
S = n/2[2a + (n − 1) × d] ....(2)
where S = sum, a = first term, n = number of terms, d = common difference.
Coming to the numerical, we are given,
a = 7, d = 14 - 7 = 7, tn = 98.
So, we find the number of terms first using the (1),
tn = a + (n – 1) × d
⇒ 98 = 7 + (n – 1) × 7
⇒ n - 1 = 91 / 7
⇒ n = 13 + 1
= 14
Now, we need to find the sum of the terms using (2),
Putting respective values in (2), we get;
S = n/2[2a + (n − 1) × d]
⇒ S = 14/2 [ 2 × 7 + (14 − 1) × 7 ]
⇒ S = 7 × [ 14 + 91 ]
⇒ S = 735
Hence, the sum of all the natural numbers from 1 to 100 that are divisible by 7 is 735.
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