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

Solution:

(i) $(b+c)(b+d)=(c+a)(c+a)$

To prove: $(b+c)(b+d)=(c+a)(c+a)$

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)

Taking LHS = (b + c)(b + d)

$=b^{2}+b d+b c+c d$

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

$=a c+c^{2}+a d+c d$

$=c(a+c)+d(a+c)$

$=(a+c)(c+d)$

Hence Proved

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

To prove: $\frac{a b-c d}{b^{2}-c^{2}}=\frac{a+c}{b}$

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

Proof: When $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ 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)

$\Rightarrow \mathrm{c}^{2}=\mathrm{bd}$

$\Rightarrow d=\frac{c^{2}}{b}$ … (iii)

Taking LHS = $\frac{a b-c d}{b^{2}-c^{2}}$

$=\frac{a b-c \frac{c^{2}}{b}}{b^{2}-c^{2}}$ [From eqn. (iii)]

$=\frac{a b-\frac{c^{3}}{b}}{b^{2}-c^{2}}$

$=\frac{\frac{a b^{2}-c^{3}}{b}}{b^{2}-c^{2}}$

$=\frac{a b^{2}-c^{3}}{b\left(b^{2}-c^{2}\right)}$

$=\frac{a^{2} c-c^{3}}{b a c-b c^{2}}$ [From eqn. (ii)]

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

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

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

$=\frac{(a+c)}{b}$

= RHS

Hence Proved

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

To prove: $(a+b+c+d)^{2}=(a+b)^{2}+2(b+c)^{2}+(c+d)^{2}$

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

Proof: When $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ 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)

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

$\Rightarrow(a+b+c+d)(a+b+c+d)$

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

On rearranging

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

On rearranging

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

On rearranging

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

On rearranging

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

Using eqn. (i)

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

Using eqn. (ii)

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

Using eqn. (iii)

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

On rearranging

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

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

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

= RHS

Hence proved