Answer:
Solution given:
first term [a]=100
last term[l]=300
common difference [d]=5
second term[b]=100+5=105
now
we have
l=a+(n-1)d
substituting value
300=100+(n-1)5
300-100=(n-1)5
200/5=n-1
40=n-1
n=40+1
there are 41 multiple in between 100 and 300/including it but no including its 41-2=39 multiple
Explanation:
Consider AP:
As multiples of 5 between 100 and 300
The first term is 105, the common difference is 5, the last term is 295.
Use the nth term formula:
4.7m questions
6.2m answers