Question

find the sum of all the 3-digit no. which are multiples of 11​

Answer 1

Answer:

The sum of all the 3 digit no. is 44,550

Answer 2

Answer:-

• The sequence of all the three - digit numbers is 100 , 101 , 102...999.

• The sequence with multiples of 11 is 110 , 121 , 132..990.

If we assume that this sequence is in AP,

• a (first term) = 110

• d (Common difference) = t(2) - t(1) = 121 - 110 = 11

• nth term (an) or (l) = 990.

We know that,

nth term of an AP = a + (n - 1)d

→ 990 = 110 + (n - 1)(11)

→ 990 = 110 + 11n - 11

→ 990 - 110 + 11 = 11n

→ 11n = 891

→ n = 891/11

→ n = 81.

And,

Sum of first n terms of an AP = n/2 * (a + l)

→ S(81) = 81/2 * (110 + 990)

→ S(81) = 81/2 * (1100)

→ S(81) = 81(550)

→ S(81) = 44,550.

Therefore, the sum of all the three digit numbers which are multiples of 11 is 44,550.