Question

Find the missing term in the series: \[ \frac{1}{2},\;\frac{9}{2},\;\frac{25}{2},\;\frac{49}{2},\;\frac{81}{2},\;\_\_\_,\;\frac{169}{2},\;\frac{225}{2},\;\frac{289}{2} \]

• A. \(\frac{111}{2}\)
• B. \(\frac{121}{2}\)
• C. \(\frac{131}{2}\)
• D. \(\frac{141}{2}\)

Hint:

In fraction series, check numerator and denominator separately. Often only one part follows the pattern.

Answer 1

The Correct Option is B

Concept: Observe the numerators carefully. They are perfect squares of consecutive odd numbers.

Step 1: Write numerators separately.
\[ 1,\;9,\;25,\;49,\;81,\;\_\_,\;169,\;225,\;289 \]

Step 2: Express them as squares.
\[ 1=1^2 \] \[ 9=3^2 \] \[ 25=5^2 \] \[ 49=7^2 \] \[ 81=9^2 \] \[ 169=13^2 \] \[ 225=15^2 \] \[ 289=17^2 \] Pattern: \[ 1^2,\;3^2,\;5^2,\;7^2,\;9^2,\;11^2,\;13^2,\;15^2,\;17^2 \] So missing term: \[ 11^2=121 \] Thus: \[ \frac{121}{2} \] Hence, the required answer is: \[ \boxed{\frac{121}{2}} \]