Question

The sum of all natural numbers from 1 to 200 which are divisible by 5

Answer 1

Step-by-step explanation:

The common difference between the sum of all natural numbersbfrom 1 to 200...which are divisible by 5 are°}

First term 5

Last term 200

common difference (d) is 5

and number of terms(n) is the 40

so

sum will be

sum =n÷2 (2(a)+(n-1)d)

sum =n÷2 (2(a)+(n-1)d)=40÷2 (2 (5)+(40-1)5)

sum =n÷2 (2(a)+(n-1)d)=40÷2 (2 (5)+(40-1)5)=20 (10+195)

sum =n÷2 (2(a)+(n-1)d)=40÷2 (2 (5)+(40-1)5)=20 (10+195)=20 (205)

sum =n÷2 (2(a)+(n-1)d)=40÷2 (2 (5)+(40-1)5)=20 (10+195)=20 (205)=4100..

Answer 2

Answer:

A.p =5,10,15,.......200

a=5, d=10-5=5 , n=?

an=a+(n-1)d

200=5+(n-1)5

200-5=5(n-1)

195/5=n--1

39=n-1

n=39+1=40

sn=n/2[2a+(n-1)d]

sn=40/2[2.5+(40-1)5]

=20[10+195]

=20×205

= 4100