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# Find the sum of all-natural numbers lying between 100 and 1000, which are multiples of 5

**Solution:**

The natural numbers lying between 100 and 1000 ,

which are multiples of 5 are

105, 110, 115, ...., 995

Here, first term, a = 105

Common difference, d = 5

Last term l = a_{n} = 995

Therefore,

a_{n} = a + (n - 1) d

105 + (n - 1)5 = 995

(n - 1)5 = 995 - 105

(n - 1) = 890/5

n = 178 + 1

n = 179

Now,

S_{179} = 179/2 [2(105) + (179 - 1) x 5]

= 179/2 [2(105) + (178)(5)]

= 179 [105 + 89(5)]

= 179 [105 + 445]

= 179 x 550

S_{179 }= 98450

Thus, the sum of all-natural numbers lying between 100 and 1000, which are multiples of 5 is 98450

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.2 Question 2

## Find the sum of all-natural numbers lying between 100 and 1000, which are multiples of 5

**Summary:**

We had to find the sum of all natural numbers lying between 100 and 1000 which are also a multiple of 5. The sum came out to be 98450

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