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

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

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Numbers divisible by 5 from 100 to 1000 are 105,110,.............995

This sequence is an A.P.

Here , first term =a =105

common difference = 5.

We know , a_n = a+(n-1)d

              995 = 105+(n-1)5

      \Rightarrow \, \, 890 = (n-1)5

    \Rightarrow \, \, 178 = (n-1)

 \Rightarrow \, \, n=178+1=179

 

S_n = \frac{n}{2}[2a+(n-1)d]

       = \frac{179}{2}[2(105)+(179-1)5]

     = \frac{179}{2}[2(105)+178(5)]

      = 179\times 550

     = 98450

The  sum of numbers divisible by 5 from 100 to 1000 is 98450.

Posted by

seema garhwal

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