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Question
If a, b, c, d are in G.P., prove that:
(a2 + b2 + c2), (ab + bc + cd), (b2 + c2 + d2) are in G.P.
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Solution
a, b, c and d are in G.P.
\[\therefore b^2 = ac\]
\[ad = bc \]
\[ c^2 = bd\] .......(1)
\[\left( ab + bc + cd \right)^2 = \left( ab \right)^2 + \left( bc \right)^2 + \left( cd \right)^2 + 2a b^2 c + 2b c^2 d + 2abcd\]
\[ \Rightarrow \left( ab + bc + cd \right)^2 = a^2 b^2 + b^2 c^2 + c^2 d^2 + a b^2 c + a b^2 c + b c^2 d + b c^2 d + abcd + abcd\]
\[ \Rightarrow \left( ab + bc + cd \right)^2 = a^2 b^2 + b^2 c^2 + c^2 d^2 + b^2 \left( b^2 \right) + ac\left( ac \right) + c^2 \left( c^2 \right) + bd\left( bd \right) + bc\left( bc \right) + ad\left( ad \right) \left[ \text { Using } (1) \right]\]
\[ \Rightarrow \left( ab + bc + cd \right)^2 = a^2 b^2 + a^2 c^2 + a^2 d^2 + b^4 + b^2 c^2 + b^2 d^2 + c^2 b^2 + c^4 + c^2 d^2 \]
\[ \Rightarrow \left( ab + bc + cd \right)^2 = a^2 \left( b^2 + c^2 + d^2 \right) + b^2 \left( b^2 + c^2 + d^2 \right) + c^2 \left( b^2 + c^2 + d^2 \right)\]
\[ \Rightarrow \left( ab + bc + cd \right)^2 = \left( b^2 + c^2 + d^2 \right)\left( a^2 + b^2 + c^2 \right)\]
\[\text { Therefore, }\left( a^2 + b^2 + c^2 \right), \left( ab + bc + cd \right) \text{ and }\left( b^2 + c^2 + d^2 \right) \text {are also in G . P } .\]
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