Question

Find the sum of all 3 – digit natural numbers which are divisible by 13.

Answer 1

We know that the sum of terms for an A.P is given by

S_n = \(\frac{n}{2}\)[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 are 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.

And, these terms form an A.P. with the common difference of 13.

Here, first term (a) = 104 and the last term (l) = 988

Common difference (d) = 13

Finding the number of terms in the A.P. by, a_n = a + (n − 1)d

We have,

988 = 104 + (n – 1)13

⟹ 988 = 104 + 13n -13

⟹ 988 = 91 + 13n

⟹ 13n = 897

⟹ n = 69

Now, using the formula for the sum of n terms, we get

S₆₉  =\(\frac{ 69}{2}\)[2(104) + (69 − 1)13]

= \(\frac{ 69}{2}\)[208 + 884]

= \(\frac{ 69}{2}\)[1092]

= 69(546)

= 37674

Hence, the sum of all 3 digit multiples of 13 is 37674.