# If a, b, c, d are in GP, then prove that

Question:

If a, b, c, d are in GP, then prove that

$\frac{1}{\left(a^{2}+b^{2}\right)}, \frac{1}{\left(b^{2}+c^{2}\right)}, \frac{1}{\left(c^{2}+d^{2}\right)}$ are in GP

Solution:

To prove: $\frac{1}{\left(a^{2}+b^{2}\right)}, \frac{1}{\left(b^{2}+c^{2}\right)}, \frac{1}{\left(c^{2}+d^{2}\right)}$ are in GP.

Given: $a, b, c, d$ are in GP

Proof: When $a, b, c, d$ are in GP then

$\Rightarrow \frac{b}{a}=\frac{c}{b}=\frac{d}{c}$

From the above, we can have the following conclusion

$\Rightarrow \mathrm{bc}=\mathrm{ad} \ldots$ (i)

$\Rightarrow \mathrm{b}^{2}=\mathrm{ac} \ldots$ (ii)

$\Rightarrow \mathrm{c}^{2}=\mathrm{bd} \ldots$ (iii)

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

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

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

From eqn. (i) , (ii) and (iii)

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

$=\frac{1}{b^{4}+2 b^{2} c^{2}+c^{4}}$

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

From the above equation, we can say that $\frac{1}{\left(a^{2}+b^{2}\right)}, \frac{1}{\left(b^{2}+c^{2}\right)}, \frac{1}{\left(c^{2}+d^{2}\right)}$

are in GP.