Question

Find the missing term in the series: \[ -\frac12,\;-\frac16,\;\frac16,\;\frac12,\;\frac56,\;\_\_\_,\;2\frac32,\;1\frac56,\;2\frac16 \]

• A. \(1\frac{1}{11}\)
• B. \(1\frac{1}{8}\)
• C. \(1\frac{1}{6}\)
• D. \(1\)

Hint:

When fractions are involved, convert everything into improper fractions first. It makes arithmetic progression patterns easier to spot.

Answer 1

The Correct Option is C

Concept: Observe the pattern of addition. Convert mixed fractions into improper fractions.

Step 1: Write the sequence clearly.
\[ -\frac12,\;-\frac16,\;\frac16,\;\frac12,\;\frac56,\;\_\_,\;\frac32,\;\frac{11}{6},\;\frac{13}{6} \]

Step 2: Find the differences.
\[ -\frac16-\left(-\frac12\right)=\frac13 \] \[ \frac16-\left(-\frac16\right)=\frac13 \] \[ \frac12-\frac16=\frac13 \] \[ \frac56-\frac12=\frac13 \] So the pattern is: \[ +\frac13 \] continuously.

Step 3: Find the missing term.
After: \[ \frac56 \] add: \[ \frac13 \] \[ \frac56+\frac13=\frac56+\frac26=\frac76 \] Convert: \[ \frac76=1\frac16 \] Check further: \[ \frac76+\frac13=\frac96=\frac32 \] Matches. Thus, the missing term is: \[ \boxed{1\frac16} \]