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Question

Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

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Solution

The integers 2,4,6,,100 lie between 1to 100 and are divisible by 2.

This shows that the series is in A.P. as,

The first term is a=2.

The common difference is d=2.

The last term is a n =100.

The n th term of the series is,

a n =a+( n1 )d 100=2+( n1 )2 50=1+n1 n=50

So, sum of the 50 th term is,

S n = 50 2 ( 2( 2 )+( 501 )( 2 ) ) = 50 2 ( 4+49×2 ) =25( 102 ) =2550

Therefore, 2550 is the required sum.

The integers 5,10,15,,100 lie between 1to 100 and are divisible by 5.

This shows that the series is in A.P. as,

The first term is a=5.

The common difference is d=5.

The last term is a n =100.

The n th term of the series is,

a n =a+( n1 )d 100=5+( n1 )5 5n=100 n=20

So, sum of the 20 th term is,

S n = 20 2 [ 2( 5 )+( 201 )5 ] =10[ 10+( 19 )5 ] =10×105 =1050

Therefore, 1050 is the required sum.

The integers 10,20,30,,100 are divisible by both 2 and 5.

This shows that the series is in A.P. as,

The first term is a=10.

The common difference is d=10.

The last term is a n =100.

The n th term of the series is,

100=10+( n1 )10 100=10n n=10

So, sum of the 10 th term is,

S n = 10 2 [ 2( 10 )+( 101 )10 ] =5[ 20+( 9 )10 ] =5×110 =550

The total sum is,

S n =2550+1050550 =3050

Therefore, 3050 is the sum of the integers from 1 to 100 which are divisible by 2 or 5.


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