The smallest number multiple by 13 is 13, The largest number that is smaller than 100 but multiple by 13 is 91.

Now the multiples of 13 lying between 13 and 91 form an arithmetic progression with a common difference of 13.

The nth term of an arithmetic progression is Tn = a + (n-1)*d.

13 + (n-1)*13 = 91

=> n-1 = (91 – 13) /13 =

=> n = 7

The sum of the first n terms of an arithmetic progression is given b (n/2)*(2a + (n-1)*d)

=> (7/2)*(26 + 6*13)

=> (7/2)*104

=> 364

**Therefore the sum of all the numbers divisible by 13 and less than 100 is 364.**

To find the sum of all numbers divisible by 13 and smaller than 100.

Let a1 be the first number divisible by 13.

Then a1 = 13.

Let 2nd number divisible by 13 be a2 = 13*2 = 26.

In a similar way the nth number an divisible by 13 is an = 13*n.

Since an is the last term, an = 13n < 100. So n = 7, as 13*7 = 91. and 13*8 =104 > 100.

Therefore the sum of all the numbers divisible by 13 and less than 100 is S8 = a1+a2+a3+..a7 = 13*1+13*2+13*3...13*7.

S8 = 13(1+2+3+...7).

S8 = 13(n+10n/2.

S8 = 13(7+1)7/2 , as 1+2+3+...n.

S8 = 13*8*7/2 = 364.

Therefore the required sum = 364.