Question

find the sum of all 3-digit natural numbers which are divisible by 13.

Answer 1

Sum of all 3-digit natural numbers, which are divisible by 13 = 37674

Solution:

The sum of 3-digit number between 100 and 999 that are divisible by 13 can be found out by arithmetic sum i.e.

First 3-digit number that is divided by 13 is 104

Greatest 3-digit number that is divided by 13 is 988

Formula for the sum of Arithmetic progression is $\bold{\frac{n}{2}(a+l)}$  with “a” being the value of the first number of the series and “l” being the last.

Therefore, a = 104 and l = 988

Value of n depends on the larger number which is divisible by 13 that is 988 by 13 is 76, whereas the number 104 divided by 13 is 8, so the number of terms is 76 – 13 = 69

The sum is $\bold{\frac{n}{2}(a+l)=\frac{69}{2}(104+988)=37674}$.  

Answer 2

Answer:

Sum of all 3-digit natural numbers which are divisible by 13 = 37,674

Explanation:

104,117,....,988 are 3-difit natural numbers which are divisible by 13 is an A.P

First term (a) = 104

Common difference (d)=a2-a1

d = 117-104 = 13 ----(1)

______________________

We know that,

$\boxed { n^{th}\: term (a_{n})=a+(n-1)d }$

______________________

Here ,

$a_{n}=988$

$\implies 104+(n-1)13=988$

Divide each term by 13 , we get

$\implies 8+n-1=76$

$\implies n+7=76$

$\implies n = 76-7$

$\implies n = 69$---(1)

_________________________

we know that,

$\boxed {Sum \:of\: n\: term =S_{n}\\=\frac{n}{2}(a+a_{n})}$

________________________

$S_{69}=\frac{69}{2}(104+988)$

=$\frac{69\times 1092}{2}$

=$69 \times 546$

=$37674$

Therefore,

Sum of all 3-digit natural numbers which are divisible by 13 = 37,674

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