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Question 2 (iii)
Find the sum of those integers from 1 to 500 which are multiples of 2 or 5.

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Solution

Since, multiples of 2 or 5 = Multiple of 2 + Multiple of 5 – Multiple of LCM (2,5) i.e. 10.
Multiples of 2 or 5 from 1 to 500
= List of multiple of 2 from 1 to 500 + List of multiple of 5 from 1 to 500 - List of multiple of 10 from 1 to 500
= (2, 4, 6, ... , 500) + (5, 10, 15, ... ,500) - (10, 20, ... ,500)

Each of the above series form an AP.

Let the number of terms in first series be n1.
So, 500=2+(n11)2 [nth term of an AP =a+(n1)d]]
498=(n11)2
n11=249
n1=250

Let the number of terms in the second list be n2.
So, 500=5+(n21)5495=(n21)5
99=(n21)n2=100

Let the number of terms in the third list be n3.
So,500=10+(n31)10490=(n31)1049010=n31n3=50

Sum of multiples of 2 or 5 from 1 to 500
= Sum of (2,4,6 … 500) + Sum of (5, 10, ….500) – Sum of (10,20,….500)
=n12[2+500]+n22[5+500]n32[10+500][Sn=n2(a+l)]
=2502×502+1002×505502×510
=250×251+505×5025×510=62750+2525012750
=8800012750=75250

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