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A day full of math games & activities. Find one near you.
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 = an = 995
Therefore,
an = a + (n - 1) d
105 + (n - 1)5 = 995
(n - 1)5 = 995 - 105
(n - 1) = 890/5
n = 178 + 1
n = 179
Now,
S179 = 179/2 [2(105) + (179 - 1) x 5]
= 179/2 [2(105) + (178)(5)]
= 179 [105 + 89(5)]
= 179 [105 + 445]
= 179 x 550
S179 = 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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