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

Question:

If abcd are in G.P., prove that:

(i) (a2 + b2), (b2 + c2), (c2 + d2) are in G.P.

(ii) (a2 − b2), (b2 − c2), (c2 − d2) are in G.P.

(iii) $\frac{1}{a^{2}+b^{2}}, \frac{1}{b^{2}-c^{2}}, \frac{1}{c^{2}+d^{2}}$ are in G.P.

(iv) (a2 + b2 + c2), (ab + bc + cd), (b2 + c2 + d2) are in G.P.

Solution:

a, b, c and d are in G.P.

$\therefore b^{2}=a c$    ...(1)

$a d=b c$

$c^{2}=b d$

(i) $\left(b^{2}+c^{2}\right)^{2}=\left(b^{2}\right)^{2}+2 b^{2} c^{2}+\left(c^{2}\right)^{2}$

$\Rightarrow\left(b^{2}+c^{2}\right)^{2}=(a c)^{2}+b^{2} c^{2}+b^{2} c^{2}+(b d)^{2} \quad[$ Using $(1)]$

$\Rightarrow\left(b^{2}+c^{2}\right)^{2}=a^{2} c^{2}+a^{2} d^{2}+b^{2} c^{2}+b^{2} d^{2} \quad[\mathrm{Using}(1)]$

$\Rightarrow\left(b^{2}+c^{2}\right)^{2}=a^{2}\left(c^{2}+d^{2}\right)+b^{2}\left(c^{2}+d^{2}\right)$

$\Rightarrow\left(b^{2}+c^{2}\right)^{2}=\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)$

Therefore, $\left(a^{2}+b^{2}\right),\left(c^{2}+d^{2}\right)$ and $\left(b^{2}+c^{2}\right)$ are also in G.P.

(ii) $\left(b^{2}-c^{2}\right)^{2}=\left(b^{2}\right)^{2}-2 b^{2} c^{2}+\left(c^{2}\right)^{2}$

$\Rightarrow\left(b^{2}-c^{2}\right)^{2}=(a c)^{2}-b^{2} c^{2}-b^{2} c^{2}+(b d)^{2}$         [Using (1)]

$\Rightarrow\left(b^{2}-c^{2}\right)^{2}=a^{2} c^{2}-b^{2} c^{2}-a^{2} d^{2}+b^{2} d^{2}$         [Using (1)]

$\Rightarrow\left(b^{2}-c^{2}\right)^{2}=c^{2}\left(a^{2}-b^{2}\right)-d^{2}\left(a^{2}-b^{2}\right)$

$\Rightarrow\left(b^{2}-c^{2}\right)^{2}=\left(a^{2}-b^{2}\right)\left(c^{2}-d^{2}\right)$

Therefore, $\left(a^{2}-b^{2}\right),\left(b^{2}-c^{2}\right)$ and $\left(c^{2}-d^{2}\right)$ are also in G.P.

$($ iii $)\left(\frac{1}{b^{2}+c^{2}}\right)^{2}=\left(\frac{1}{b^{2}}\right)^{2}+\frac{2}{b^{2} c^{2}}+\left(\frac{1}{c^{2}}\right)^{2}$

$\Rightarrow\left(\frac{1}{b^{2}+c^{2}}\right)^{2}=\left(\frac{1}{a c}\right)^{2}+\frac{1}{b^{2} c^{2}}+\frac{1}{b^{2} c^{2}}+\left(\frac{1}{b d}\right)^{2} \quad[$ Using $(1)]$

$\Rightarrow\left(\frac{1}{b^{2}+c^{2}}\right)^{2}=\frac{1}{a^{2} c^{2}}+\frac{1}{a^{2} d^{2}}+\frac{1}{b^{2} c^{2}}+\frac{1}{b^{2} d^{2}} \quad[\operatorname{Using}(1)]$

$\Rightarrow\left(\frac{1}{b^{2}+c^{2}}\right)^{2}=\frac{1}{a^{2}}\left(\frac{1}{c^{2}}+\frac{1}{d^{2}}\right)+\frac{1}{b^{2}}\left(\frac{1}{c^{2}}+\frac{1}{d^{2}}\right)$

$\Rightarrow\left(\frac{1}{b^{2}+c^{2}}\right)^{2}=\left(\frac{1}{a^{2}+b^{2}}\right)\left(\frac{1}{c^{2}}+\frac{1}{d^{2}}\right)$

Therefore, $\left(\frac{1}{b^{2}+c^{2}}\right),\left(\frac{1}{c^{2}+d^{2}}\right)$ and $\left(\frac{1}{b^{2}+c^{2}}\right)$ are also in G.P.

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

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

$\Rightarrow(a b+b c+c d)^{2}=a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+b^{2}\left(b^{2}\right)+a c(a c)+c^{2}\left(c^{2}\right)+b d(b d)+b c(b c)+a d(a d)$            [Using (1)]

$\Rightarrow(a b+b c+c d)^{2}=a^{2} b^{2}+a^{2} c^{2}+a^{2} d^{2}+b^{4}+b^{2} c^{2}+b^{2} d^{2}+c^{2} b^{2}+c^{4}+c^{2} d^{2}$

$\Rightarrow(a b+b c+c d)^{2}=a^{2}\left(b^{2}+c^{2}+d^{2}\right)+b^{2}\left(b^{2}+c^{2}+d^{2}\right)+c^{2}\left(b^{2}+c^{2}+d^{2}\right)$

$\Rightarrow(a b+b c+c d)^{2}=\left(b^{2}+c^{2}+d^{2}\right)\left(a^{2}+b^{2}+c^{2}\right)$

Therefore, $\left(a^{2}+b^{2}+c^{2}\right),(a b+b c+c d)$ and $\left(b^{2}+c^{2}+d^{2}\right)$ are also in G.P.

 

 

 

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