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

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)

Answer 2

Answer: The missing number in the sequence is
`(7)/(6)`

Explanation:

Since we have given that

`(1)/(6),(1)/(2),(5)/(6),----------,(11)/(2)`

First term = a=
`(1)/(6)`

Common difference = d is given by

`d=a_2-a_1\\\\(1)/(2)-(1)/(6)\\\\=(6-2)/(12)\\\\=(4)/(12)=(1)/(3)\\\\Similarly,\\d=a_3-a_2\\\\=(5)/(6)-(1)/(2)\\\\=(10-6)/(12)\\\\=(4)/(12)\\\\=(1)/(3)`

Therefore, it forms an arithmetic sequence.

Since,
`a_4` is missing,

So,

`a_4=a+3d\\\\a_4=(1)/(6)+3* (1)/(3)\\\\a_4=(1)/(6)+1\\\\a_4=(1+6)/(6)\\\\a_4=(7)/(6)`

Hence, the missing number in the sequence is
`(7)/(6)`