Question

Joshua uses a rule to write the following sequence of numbers 1/6,1/2,5/6,----------,11/2 What rule did Joshua use? What is the missing number in the sequence?

Answer 1

Answer: The missing number in the sequence is $\frac{7}{6}$

Step-by-step explanation:

Since we have given that

$\frac{1}{6},\frac{1}{2},\frac{5}{6},----------,\frac{11}{2}$

First term = a= $\frac{1}{6}$

Common difference = d is given by

$d=a_2-a_1\\\\\frac{1}{2}-\frac{1}{6}\\\\=\frac{6-2}{12}\\\\=\frac{4}{12}=\frac{1}{3}\\\\Similarly,\\d=a_3-a_2\\\\=\frac{5}{6}-\frac{1}{2}\\\\=\frac{10-6}{12}\\\\=\frac{4}{12}\\\\=\frac{1}{3}$

Therefore, it forms an arithmetic sequence.

Since, $a_4$ is missing,

So,

$a_4=a+3d\\\\a_4=\frac{1}{6}+3\times \frac{1}{3}\\\\a_4=\frac{1}{6}+1\\\\a_4=\frac{1+6}{6}\\\\a_4=\frac{7}{6}$

Hence, the missing number in the sequence is $\frac{7}{6}$

Answer 2

This is an arithmetic progression at a rate of 2/6. 

So 1/6 + 2/6 = 3/6 (1/2)
3/6 + 2/6 = 5/6
5/6 + 2/6 = 7/6
....
29/6 + 2/6 = 31/6 
31/6 +2/6 = 33/6 (11/2)