If $2, b, c, 23$ are in G. P. then ${(b - c)}^{2} + {(c - 2)}^{2} + {(23 - b)}^{2} =$

$625$

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$525$

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$441$

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$442$

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Solution

The correct option is C
$441$

Explanation for the correct options:
Finding the value.
Given Data: $2, b, c, 23$ are in G. P
Let $r$ be the common ratio of GP.
So $b = 2 r, c = 2 r^{2}, 23 = 2 r^{3}$
Now, ${(b - c)}^{2} + {(c - 2)}^{2} + {(23 - b)}^{2}$
$= {(2 r - 2 r^{2})}^{2} + {(2 r^{2} - 2)}^{2} + {(23 - 2 r)}^{2}$
$= 4 {(r - r^{2})}^{2} + 4 {(r^{2} - 1)}^{2} + {(23 - 2 r)}^{2}$
$= 4 (r^{2} - 2 r^{3} + r^{4} + r^{4} - 2 r^{2} + 1) + (529 - 92 r + 4 r^{2})$
$= 8 r^{4} - 8 r^{3} - 4 r^{2} + 4 + 529 - 92 r + 4 r^{2}$
$= 8 r (\frac{23}{2}) - 8 (\frac{23}{2}) - 92 r + 533$
$= 92 r - 92 - 92 r + 533$
$= 441$
Hence, Option(C) is correct.

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