Q. 24.5( 109 Votes )
Find the sum of all-natural numbers lying between 100 and 1000, which are multiples of 5.
The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.
Let the first term be ‘a’ and common difference ‘d’.
Let n be the total number of terms in the series.
Here, first term, a = 105
Common difference, d = 5
If l denotes the last term of the series
Then, l = a + (n – 1) × d
Here, l = 995
⇒ 995 = 105 + (n – 1) × 5
⇒ 995 – 105 = (n – 1) × 5
⇒ 890/5 = n – 1
⇒ 178 + 1 = n
∴ n = 179
Sum of A.P. =
⇒ Sn = 179 × 550 = 98,450.
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