If a, b, c, d are in G.P., prove that: (a^2+b^2+c^2 ), (ab+bc+cd ), (b^2+c^2+d^2 ) are in G.P.

a, b, c, d are in G.P. ^2=ac ad=bc c^2=bd (1) (ab+bc+cd)^2= (ab )^2+ (bc )^2+ (cd )^2+2ab^2c+2bc^2d+2abcd (ab+bc+cd)^2=a^2b^2+b^2c^2+c^2d^2+ab^2c+bc^2d+abcd+abcd (ab+bc+cd )^2=a^2b^2+b^2c^2+c^2d^2+b^2 (b^2 ) +ac(ac)+c^2(c)^2+bd(bd)+bc(bc)+ad(ad) (ab+bc+cd)^2=a^2b^2+b^2c^2+a^2d^2+b^4+b^2c^2+b^2d^2+c^2b^2+c^4+c^2d^2 (ab+bc+cd)^2=a^2 (b^2+c^2+d^2 )+b^2 (b^2+c^2+d^2 )+c^2 (b^2+c^2+d^2 ) (ab+bc+cd)^2= (b^2+c^2+d^2 ) (a^2+b^2+c^2 ) (a^2+b^2+c^2 ),(ab+bc+cd) and (b^2+c^2+d^2 ) are also in G.P.

If a, b, c, d are in G.P., prove that: (a^2+b^2+c^2 ), (ab+bc+cd …

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Question

If a, b, c, d are in G.P., prove that:
$\big(\text{a}^2+\text{b}^2+\text{c}^2\big),\big(\text{ab}+\text{bc}+\text{cd}\big),\big(\text{b}^2+\text{c}^2+\text{d}^2\big)\text{ are in G.P.}$

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Answer

a, b, c, d are in G.P.
$\therefore\text{b}^2=\text{ac}$
$\text{ad}=\text{bc}$
$\text{c}^2=\text{bd}\cdots(1)$
$(\text{ab}+\text{bc}+\text{cd})^2=\big(\text{ab}\big)^2+\big(\text{bc}\big)^2+\big(\text{cd}\big)^2+2\text{ab}^2\text{c}+2\text{bc}^2\text{d}+2\text{abcd}$
$(\text{ab}+\text{bc}+\text{cd})^2=\text{a}^2\text{b}^2+\text{b}^2\text{c}^2+\text{c}^2\text{d}^2+\text{ab}^2\text{c}+\text{bc}^2\text{d}+\text{abcd}+\text{abcd}$
$\Rightarrow\big(\text{ab}+\text{bc}+\text{cd}\big)^2=\text{a}^2\text{b}^2+\text{b}^2\text{c}^2+\text{c}^2\text{d}^2+\text{b}^2\big(\text{b}^2\big)\\+\text{ac}(\text{ac})+\text{c}^2(\text{c})^2+\text{bd}(\text{bd})+\text{bc}(\text{bc})+\text{ad}(\text{ad})$
$$$\Rightarrow(\text{ab}+\text{bc}+\text{cd})^2=\text{a}^2\text{b}^2+\text{b}^2\text{c}^2+\text{a}^2\text{d}^2+\text{b}^4+\text{b}^2\text{c}^2+\text{b}^2\text{d}^2+\text{c}^2\text{b}^2+\text{c}^4+\text{c}^2\text{d}^2$
$\Rightarrow(\text{ab}+\text{bc}+\text{cd})^2=\text{a}^2\big(\text{b}^2+\text{c}^2+\text{d}^2\big)+\text{b}^2\big(\text{b}^2+\text{c}^2+\text{d}^2\big)+\text{c}^2\big(\text{b}^2+\text{c}^2+\text{d}^2\big)$
$\Rightarrow(\text{ab}+\text{bc}+\text{cd})^2=\big(\text{b}^2+\text{c}^2+\text{d}^2\big)\big(\text{a}^2+\text{b}^2+\text{c}^2\big)$
$\therefore\big(\text{a}^2+\text{b}^2+\text{c}^2\big),(\text{ab}+\text{bc}+\text{cd})\text{ and }\big(\text{b}^2+\text{c}^2+\text{d}^2\big)\text{ are also in G.P.}$