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The sum of integers from 1 to 100 that are divisible by 2 or 5 is A. 3000 B. 3050 C. 3600 D. 3250

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Numbers that are divisible by 2 and 5 are 2,4,5,6,8,10.............92,94,95,96,98,100. 

A1 = 2,4,6,8,10,...100 are in AP with common difference(d1) = 2. & A2 = 5,10,15,...100 are also in AP with common difference (d2 =5).

and A3 = 10,20,30......100 are  also in AP with common diff. (d3) = 10.
n term of A1 A2 & A3 can be obtain by using formula, [a + (n-1)d = last term] , 

so total no. of terms in A1 , A2 & A3 are 50, 20, 10 resp.

S1 = N1/2{2a +49d1}  ,  S2 = N2/2{2a +19d2} & S3 = N3/2{2a + 9d3}

S1 = 50/2{2*2 + 49*2}  = 25{4+98} =2550

S2 = 20/2{2*5 + 19*5} = 10{10+95} =1050

S3 = 10/2{2*10 + 9*10} = 5{20+90} = 550

total sum = S1+ S2 -S3 = 3050 
correct : Option (B)

 

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manish

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