Question

find the sum of all the numbers lying between hundred and thousand that are multiple of 2 and 5​

Answer 1

Answer:-

• The sequence of numbers which are multiples of 2 between 100 and 1000 is 102,104,106... 998.

• The sequence of numbers which are multiples of 5 between 100 and 1000 is 105,110,115...995.

• Common terms from both the sequences are 110,120,....990.

If we assume that this sequence is in AP then,

• a (first term) = 110

• d (Common difference) = t(2) - t(1) = 120 - 110 = 10.

• nth term = 990.

We know that,

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

→ 990 = 110 + (n - 1)10

→ 990 = 110 + 10n - 10

→ 990 - 110 + 10 = 10n

→ 890 = 10n

→ n = 890/10

→ n = 89

We know that,

Sum of first n terms of an AP = n/2* [ 2a + (n - 1)d]

→ S(89) = 89/2 * [ 2(110) + (89 - 1)(10)]

→ S(89) = 89/2 * [ 220 + 880 ]

→ S(89) = 89/2 * [ 1000 ]

→ S(89) = 89(500)

→ S(89) = 44,500.

Therefore, the sum of all the natural numbers lying between 100 and 1000 that are multiples of 2 and 5 is 44,500.