How many multiples of 9 are there between 100 and 2,000? Hint: an = a1 + d(n − 1), where a1 is the first term and d is the common difference. 196 211 210 212
2,000 - 100 =1,900
1,900 / 9 =211 1/9 drop the 1/9 fraction. That leaves 211 multiples of 9
The first one =108
The last one =1,998
[1,998 - 108] /9 =210 +1 =211 You have to count both the 1st and the last. Using the formula above:
1,998=108 + 9(n-1), solve for n:
n=211
