Question

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

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

Answer 1

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

$$\begin{gathered} \therefore b^2=ac \\ ad=bc \\ {c}^2=bd \end{gathered}$$ .......(1)

$$\begin{gathered} \left(\mathrm{i}\right){\left(b^2+c^2\right)}^2={\left(b^2\right)}^2+2b^2{c}^2+{\left(c^2\right)}^2 \\ \Rightarrow {\left(b^2+c^2\right)}^2={\left(ac\right)}^2+b^2{c}^2+b^2{c}^2+{\left(bd\right)}^2\left[\mathrm{Using}\left(1\right)\right] \\ \Rightarrow {\left(b^2+c^2\right)}^2=a^2{c}^2+a^2{d}^2+b^2{c}^2+b^2{d}^2\left[\mathrm{Using}\left(1\right)\right] \\ \Rightarrow {\left(b^2+c^2\right)}^2=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right) \\ \Rightarrow {\left(b^2+c^2\right)}^2=\left(a^2+b^2\right)\left(c^2+d^2\right) \\ \mathrm{Therefore},\left(a^2+b^2\right),\left(c^2+d^2\right)\mathrm{and}\left(b^2+c^2\right)\mathrm{are}\mathrm{also}\mathrm{in}\mathrm{G}.\mathrm{P}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right){\left(b^2-c^2\right)}^2={\left(b^2\right)}^2-2b^2{c}^2+{\left(c^2\right)}^2 \\ \Rightarrow {\left(b^2-c^2\right)}^2={\left(ac\right)}^2-b^2{c}^2-b^2{c}^2+{\left(bd\right)}^2\left[\mathrm{Using}\left(1\right)\right] \\ \Rightarrow {\left(b^2-c^2\right)}^2=a^2{c}^2-b^2{c}^2-a^2{d}^2+b^2{d}^2\left[\mathrm{Using}\left(1\right)\right] \\ \Rightarrow {\left(b^2-c^2\right)}^2=c^2\left(a^2-b^2\right)-d^2\left(a^2-b^2\right) \\ \Rightarrow {\left(b^2-c^2\right)}^2=\left(a^2-b^2\right)\left(c^2-d^2\right) \\ \mathrm{Therefore},\left(a^2-b^2\right),\left(b^2-c^2\right)\mathrm{and}\left(c^2-d^2\right)\mathrm{are}\mathrm{also}\mathrm{in}\mathrm{G}.\mathrm{P}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right){\left(\frac{1}{b^2+c^2}\right)}^2={\left(\frac{1}{b^2}\right)}^2+\frac{2}{b^2{c}^2}+{\left(\frac{1}{c^2}\right)}^2 \\ \Rightarrow {\left(\frac{1}{b^2+c^2}\right)}^2={\left(\frac{1}{ac}\right)}^2+\frac{1}{b^2{c}^2}+\frac{1}{b^2{c}^2}+{\left(\frac{1}{bd}\right)}^2\left[\mathrm{Using}\left(1\right)\right] \\ \Rightarrow {\left(\frac{1}{b^2+c^2}\right)}^2=\frac{1}{a^2{c}^2}+\frac{1}{a^2{d}^2}+\frac{1}{b^2{c}^2}+\frac{1}{b^2{d}^2}\left[\mathrm{Using}\left(1\right)\right] \\ \Rightarrow {\left(\frac{1}{b^2+c^2}\right)}^2=\frac{1}{a^2}\left(\frac{1}{c^2}+\frac{1}{d^2}\right)+\frac{1}{b^2}\left(\frac{1}{c^2}+\frac{1}{d^2}\right) \\ \Rightarrow {\left(\frac{1}{b^2+c^2}\right)}^2=\left(\frac{1}{a^2+b^2}\right)\left(\frac{1}{c^2}+\frac{1}{d^2}\right) \\ \mathrm{Therefore},\left(\frac{1}{b^2+c^2}\right),\left(\frac{1}{c^2+d^2}\right)\mathrm{and}\left(\frac{1}{b^2+c^2}\right)\mathrm{are}\mathrm{also}\mathrm{in}\mathrm{G}.\mathrm{P}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right){\left(ab+bc+cd\right)}^2={\left(ab\right)}^2+{\left(bc\right)}^2+{\left(cd\right)}^2+2a{b}^{2}c+2b{c}^{2}d+2abcd \\ \Rightarrow {\left(ab+bc+cd\right)}^2=a^2{b}^2+b^2{c}^2+c^2{d}^2+a{b}^{2}c+a{b}^{2}c+b{c}^{2}d+b{c}^{2}d+abcd+abcd \\ \Rightarrow {\left(ab+bc+cd\right)}^2=a^2{b}^2+b^2{c}^2+c^2{d}^2+b^2\left(b^2\right)+ac\left(ac\right)+c^2\left(c^2\right)+bd\left(bd\right)+bc\left(bc\right)+ad\left(ad\right)\left[\mathrm{Using}\left(1\right)\right] \\ \Rightarrow {\left(ab+bc+cd\right)}^2=a^2{b}^2+a^2{c}^2+a^2{d}^2+b^4+b^2{c}^2+b^2{d}^2+c^2{b}^2+c^4+c^2{d}^2 \\ \Rightarrow {\left(ab+bc+cd\right)}^2=a^2\left(b^2+c^2+d^2\right)+b^2\left(b^2+c^2+d^2\right)+c^2\left(b^2+c^2+d^2\right) \\ \Rightarrow {\left(ab+bc+cd\right)}^2=\left(b^2+c^2+d^2\right)\left(a^2+b^2+c^2\right) \\ \mathrm{Therefore},\left(a^2+b^2+c^2\right),\left(ab+bc+cd\right)\mathrm{and}\left(b^2+c^2+d^2\right)\mathrm{are}\mathrm{also}\mathrm{in}\mathrm{G}.\mathrm{P}. \end{gathered}$$