Question

Find the sum of all 3 – digit natural numbers which are multiples of 11.

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 multiples of 11.

So, we know that the first 3 digit multiple of 11 is 110 and the last 3 digit multiple of 13 is 990.

And, these terms form an A.P. with the common difference of 11.

Here, first term (a) = 110 and the last term (l) = 990

Common difference (d) = 11

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

We get,

990 = 110 + (n – 1)11

⟹ 990 = 110 + 11n -11

⟹ 990 = 99 + 11n

⟹ 11n = 891

⟹ n = 81

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

S₈₁  = \(\frac{81}{2}\)[2(110) + (81 − 1)11]

= \(\frac{81}{2}\)[220 + 880]

= \(\frac{81}{2}\)[1100]

= 81(550)

= 44550

Hence, the sum of all 3 digit multiples of 11 is 44550.