An=a1+d (n-1) A1=200 since that's the first term that can be a multiple of 5 N=? That's what we need to find An=1195 since that's the last multiple of 5 that we can use D=5 Plug in 1195=200+5 (n-1) -200 both sides 995=5 (n-1) Distribute 5 to n-1 995=5n-5 +5 both sides 1000=5n ÷5 both sides N=200 there are 200 multiples of 5 in between 199 and 1198
