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

Solution:

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

To prove: $a\left(b^{2}+c^{2}\right)=c\left(a^{2}+b^{2}\right)$

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

Formula used: When $a, b, c$ are in GP, $b^{2}=a c$

When $a, b, c$ are in GP, $b^{2}=a c$

Taking LHS $=a\left(b^{2}+c^{2}\right)$

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

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

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

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

$=\operatorname{RHS}$

Hence Proved

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

To prove: $a\left(b^{2}+c^{2}\right)=c\left(a^{2}+b^{2}\right)$

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

Formula used: When $a, b, c$ are in $G P, b^{2}=a c$

Proof: When $a, b, c$ are in GP, $b^{2}=a c$

Taking $L H S=\frac{1}{\left(a^{2}-b^{2}\right)}+\frac{1}{b^{2}}$

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

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

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

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

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

Hence Proved

(iii) $(a+2 b+2 c)(a-2 b+2 c)=a^{2}+4 c^{2}$

To prove: $(a+2 b+2 c)(a-2 b+2 c)=a^{2}+4 c^{2}$

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

Formula used: When $a, b, c$ are in GP, $b^{2}=a c$

Proof: When $a, b, c$ are in GP, $b^{2}=a c$

Taking LHS = (a + 2b + 2c)(a – 2b + 2c)

$\Rightarrow[(a+2 c)+2 b][(a+2 c)-2 b]$

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

$\Rightarrow\left[\left(a^{2}+4 a c+4 c^{2}\right)-4 b^{2}\right]$

$\Rightarrow\left[\left(a^{2}+4 a c+4 c^{2}\right)-4 b^{2}\right]\left[b^{2}=a c\right]$

$\Rightarrow\left[\left(a^{2}+4 a c+4 c^{2}-4 a c\right]\right.$

$\Rightarrow a^{2}+4 c^{2}=R H S$

Hence Proved

(iv) $a^{2} b^{2} c^{2}\left(\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}\right)=a^{3}+b^{3}+c^{3}$

To prove: $a^{2} b^{2} c^{2}\left(\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}\right)=a^{3}+b^{3}+c^{3}$

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

Formula used: When $a, b, c$ are in GP, $b^{2}=a c$

Proof: When $a, b, c$ are in GP, $b^{2}=a c$

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

$\Rightarrow\left(\frac{b^{3} c^{3}+a^{3} c^{3}+a^{3} b^{3}}{a b c}\right)$

$\Rightarrow\left(\frac{b^{2} b c^{3}+(a c)^{2} a c+a^{3} b^{2} b}{a b c}\right)$

$\Rightarrow\left(\frac{a c b c^{3}+\left(b^{2}\right)^{2} a c+a^{3} a c b}{a b c}\right)\left[b^{2}=a c\right]$

$\Rightarrow\left(\frac{a c b c^{3}+b^{3} a b c+a^{3} a c b}{a b c}\right)$

$\Rightarrow\left(\mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}\right)=\mathrm{RHS}$

Hence Proved