If $a, b, c, d$ are in $G . P$ ,prove that
$(a + b + c + d)^{2} = (a + b)^{2} + 2 (b + c)^{2} + (b + c)^{2}$

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Solution

$a, b, c, d$ are in G.P
${(a + b + c + d)}^{2} = {(a + b)}^{2} + 2 (a + b) (c + d) + {(c + d)}^{2} = {(a + b)}^{2} + 2 (a c + a d + b c + b d) + {(c + d)}^{2} = {(a + b)}^{2} + 2 (b^{2} + b c + b c + c^{2}) + {(c + d)}^{2} {(a + b + c + d)}^{2} = {(a + b)}^{2} + 2 {(b + c)}^{2} + {(c + d)}^{2} (∵ a c = b^{2}, a d = b c, b d = c^{2})$

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Similar questions

(a^{2} + b^{2}), (b^{2} + c^{2}), (c^{2} + d^{2})

If a, b, c, d are in G.P., prove that :

(i) $\frac{a b - c d}{b^{2} - c^{2}} = \frac{a + c}{b}$

(ii) $(a + b + c + d)^{2} = (a + b)^{2} + 2 (b + c)^{2} + (c + d)^{2}$

(iii) $(b + c) (b + d) = (c + d) (c + d)$

\frac{a b - c d}{b^{2} - c^{2}} = \frac{a + c}{b}

{(b - c)}^{2} + {(c - a)}^{2} + {(d - b)}^{2} = {(a - d)}^{2}

(a^{2} + b^{2} + c^{2}), (a b + b c + c d), (b^{2} + c^{2} + d^{2})

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