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

a, b, c, d are in G.P. ^2=ac ab=bc c^2=bd (1) Now, (b^2+c^2)^2= (b^2 )^22b^2c^2+ (c^2 )^2 (b^2+c^2 )^2= (ac )^2+b^2c^2+b^2c^2+ (bd )^2 [Using (1)] (b^2+c^2 )^2=a^2c^2+a^2d^2+b^2c^2+b^2d^2 [Using (1)] (b^2+c^2 )^2=a^2 (c^2+d^2 )+b^2 (c^2+d )^2 (b^2+c^2 )^2= (a^2+b^2 ) (c^2+d^2 ) (a^2+b^2 ), (c^2+d^2 ) and (b^2+c^2 ) are also in G.P.

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

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

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Answer

a, b, c, d are in G.P. $\therefore\text{b}^2=\text{ac}$ $\text{ab}=\text{bc}$ $\text{c}^2=\text{bd}\cdots(1)$ Now, $(\text{b}^2+\text{c}^2)^2=\big(\text{b}^2\big)^22\text{b}^2\text{c}^2+\big(\text{c}^2\big)^2$ $\Rightarrow\big(\text{b}^2+\text{c}^2\big)^2=\big(\text{ac}\big)^2+\text{b}^2\text{c}^2+\text{b}^2\text{c}^2+\big(\text{bd}\big)^2$ [Using (1)] $\Rightarrow\big(\text{b}^2+\text{c}^2\big)^2=\text{a}^2\text{c}^2+\text{a}^2\text{d}^2+\text{b}^2\text{c}^2+\text{b}^2\text{d}^2$ [Using (1)] $\Rightarrow\big(\text{b}^2+\text{c}^2\big)^2=\text{a}^2\big(\text{c}^2+\text{d}^2\big)+\text{b}^2\big(\text{c}^2+\text{d}\big)^2$ $\Rightarrow\big(\text{b}^2+\text{c}^2\big)^2=\big(\text{a}^2+\text{b}^2\big)\big(\text{c}^2+\text{d}^2\big)$ $\therefore\big(\text{a}^2+\text{b}^2\big),\big(\text{c}^2+\text{d}^2\big)\text{ and }\big(\text{b}^2+\text{c}^2\big)\text{ are also in G.P.}$

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