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ryzh [129]
3 years ago
7

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?
Mathematics
2 answers:
Pani-rosa [81]3 years ago
4 0

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}

Nastasia [14]3 years ago
4 0
This is an <span>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)</span>
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