Question

the sum of all natural numbers between 100 and 300, which are divisible by 5 (incl. 100 & 300), is

Answer 1

Answer:

Step-by-step explanation:

a=100 d=5 an=300

an=a+(n-1)d

300=100+(n-1)5

300-100=5n-5

200+5=5n

205/5=n

41=n

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

    41/2(2*100+(41-1)*5)

    41/2(200+40*5)

    41/2(200+200)

     41/2*400

41*200

 8200

Answer 2

Answer:

Step-by-step explanation:by using formula of an you can find n it will be equal to 41 now using formula of sn you will get the sum equal to 8200