Question

. The sum of the numbers between 101 and 305 end with 5 is;
A) 2100
B) 1300
C) 8100
D) 4000
​

Answer 1

Answer:

The list of numbers between 101 and 999 that are divisible by 2 and 5 are:

110,120,130,...990

The numbers are in A.P, with first term, a=110, common difference, d=10

Last term, a

n

=990

We know that, a

n

=a+(n−1)d

990=110+(n−1)10

⇒990–110=10n–10

⇒880+10=10n

⇒890=10n

⇒n=89

Therefore, the number of terms between 101 and 999 that are divisible by 2 and 5 are 89.

Answer 2

Answer:

4000

Step-by-step explanation:

n=greater value-lowest value /C.D +1

n=300-105/5 +1

n=40

but number ends with 5 so divide 40 by 2 so

n=20

•an=295 because number ends with 5 then put n=20,A1=105,an=295 in formula of sn then you will get 4000