Find the median of $A P$ .

\frac{\sqrt{31}}{2}

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\frac{3 \sqrt{6}}{2}

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\frac{3 \sqrt{2}}{2}

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\frac{3 \sqrt{5}}{2}

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Solution

The correct option is A $\frac{\sqrt{31}}{2}$
Median of AP $=$ $M_{a} = \frac{1}{2} \sqrt{(2 (2^{2} + 4^{2}) - 3^{2})}$
$=$ $\frac{1}{2} \sqrt{(2 (20) - 3^{2})}$
$=$ $\frac{1}{2} \sqrt{(40 - 9)}$
$=$ $\frac{1}{2} \sqrt{31}$
$=$ $\frac{\sqrt{31}}{2}$

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Question 6
Using the given pattern, find the missing numbers:
$1^{2} + 2^{2} + 2^{2} = 3^{2}$
$2^{2} + 3^{2} + 6^{2} = 7^{2}$
$3^{2} + 4^{2} + 12^{2} = 13^{2}$
$4^{2} + 5^{2} + {- --- -}^{2} = 21^{2}$
$5^{2} + {- --- -}^{2} + 30^{2} = 31^{2}$
$6^{2} + 7^{2} + - --- - 2 = - --- - 2$