Question

find the sum of all three-digit number divisible by both 2 and 5.​

Answer 1

Step-by-step explanation:

We want all three digit numbers i.e. least number is 100 and greatest number is 999.

Number must be divisible by 5. Thus,

a

1

=100

a

n

=995

d=5

nth term of A.P. is,

a

n

=a

1

+(n−1)d

∴995=100+(n−1)5

∴5(n−1)=995−100

∴5(n−1)=895

∴n−1=179

∴n=180

Sum of n terms of A.P. is,

∑a=

2

n

[2a+(n−1)d]

∴∑a=

2

180

[(2×100)+(180−1)5]

∴∑a=90[200+(179×5)]

∴∑a=90[200+895]

∴∑a=90×1095

∴∑a=98550