Question

Consider the multiples of 7 in between 100 and 500. 

a. What are the first and last numbers? 

b. How many terms are there in the sequence?

Answer 1

a. First term = 100 – 2 + 7 = 105 

Last term = 500 – 3 = 497

b. Common difference = x_n = d_n + (f - d)

497 = 7n + (105 - 7)

\(n=\frac{(497-98)}{7}=57\)

So, number of terms = 57