Question

Find the missing term in the series: \[ \frac{1}{3},\;1\frac{1}{3},\;2\frac{5}{6},\;4\frac{5}{6},\;7\frac{1}{3},\;\_\_\_,\;13\frac{5}{6},\;17\frac{5}{6},\;22\frac{1}{3} \]

• A. \(10\frac{1}{3}\)
• B. \(10\frac{2}{3}\)
• C. \(11\frac{5}{6}\)
• D. \(12\)

Hint:

For mixed-fraction series, convert them into improper fractions first. Difference patterns become much easier to identify.

Answer 1

The Correct Option is A

Concept: Observe the differences between consecutive terms.

Step 1: Convert into improper fractions.
\[ \frac13,\;\frac43,\;\frac{17}{6},\;\frac{29}{6},\;\frac{22}{3},\;\_\_,\;\frac{83}{6},\;\frac{107}{6},\;\frac{67}{3} \]

Step 2: Find the pattern in differences.
\[ \frac43-\frac13=1 \] \[ \frac{17}{6}-\frac43=\frac32 \] \[ \frac{29}{6}-\frac{17}{6}=2 \] \[ \frac{22}{3}-\frac{29}{6}=\frac52 \] So the differences are: \[ 1,\;\frac32,\;2,\;\frac52,\;3,\;\frac72,\;4,\;\frac92 \] They increase by: \[ \frac12 \] each time.

Step 3: Find the missing term.
After: \[ 7\frac13=\frac{22}{3} \] Next difference should be: \[ 3 \] So: \[ \frac{22}{3}+3=\frac{31}{3} \] Converting: \[ \frac{31}{3}=10\frac13 \] Thus, the missing term is: \[ \boxed{10\frac13} \]