Find the sum of all 3 - digit natural numbers which are divisible by 13.
Solution
In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of n terms of an A.P.,
`S_n = n/2 [2a +(n -1)d]`
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
All 3 digit natural number which is divisible by 13
So, we know that the first 3 digit multiple of 13 is 104 and the last 3 digit multiple of 13 is 988.
Also, all these terms will form an A.P. with the common difference of 13.
So here,
First term (a) = 104
Last term (l) = 988
Common difference (d) = 13
So, here the first step is to find the total number of terms. Let us take the number of terms as n.
`Now, as we know,
`a_n = a + (n - 1)d`
So, for the last term,
988 = 104 + (n - 1)13
988 = 104 + 13n - 13
988 = 91 + 13n
Further simplifying,
`n = (988 - 91)/13`
`n = 897/13`
n = 69
Now, using the formula for the sum of n terms, we get
`S_n = 69/2 [2(104) + (69 - 1)3]`
`= 69/2 [208 + (68)13]`
`= 69/2 (208 + 884)`
On further simplification, we get,
`S_n = 69/2 (1092)`
= 69 (546)
= 37674
Therefore, the sum of all the 3 digit multiples of 13 is `S_n = 37674`
