Bunuel wrote:

If each number in a sequence is three more than the previous number, and the sixth number is 32, what is the 100th number?

If each number in a sequence is three more than the previous number, and the

sixth number is 32, what is the 100th number?

The answer says -

From the sixth to the one hundredth term, there are 94 “jumps” of 3. Since 94 x 3 = 282, there is an

increase of 282 from the sixth term to the one hundredth term:

32 + 282 = 314.

Can someone please help me understand how the 94 figure is arrived at?