Question

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

Answer 1

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+( n−1 )d 100=2+( n−1 )2 50=1+n−1 n=50

So, sum of the 50 th term is,

S n = 50 2 ( 2( 2 )+( 50−1 )( 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+( n−1 )d 100=5+( n−1 )5 5n=100 n=20

So, sum of the 20 th term is,

S n = 20 2 [ 2( 5 )+( 20−1 )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+( n−1 )10 100=10n n=10

So, sum of the 10 th term is,

S n = 10 2 [ 2( 10 )+( 10−1 )10 ] =5[ 20+( 9 )10 ] =5×110 =550

The total sum is,

S n =2550+1050−550 =3050

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