Question

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

(i) $a(b^2+c^2)=c(a^2+b^2)$

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

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

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

(v) $(a+2b+2c)(a-2b+2c)=a^2+4c^2$.

(v) $(a+2b+2c)(a-2b+2c)=a^2+4c^2$.

Answer 1

(i) $a(b^2+c^2)=c(a^2+b^2)$

Since, a, b, c are in G.P.

$\therefore$ a = a, b = ar, c = $a{r}^2$

$a(b^2+c^2)=c(a^2+b^2)$

$a(a^2{r}^2+a^2{r}^4)=a{r}^2(a^2+a^2{r}^2)$

$a^3{r}^2(1+r^2)=a^3{r}^2(1+r^2)$

LHS = RHS

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

Since, a, b, c are in G.P.

$\therefore a=a,b=ar,c=a{r}^2$

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

$=a^2\times a^2{r}^2\times a^2{r}^4(\frac{1}{a^3}+\frac{1}{a^3{r}^3}+\frac{1}{a^3{r}^6})$

$=a^6{r}^6\left(\frac{r^6+r^3+1}{a^3{r}^6}\right)$

$=a^3(r^6+r^3+1)$

$=a^3+a^3{r}^3+a^3{r}^6$

$=a^3+(ar{)}^3+(a{r}^2{)}^3$

$=a^3+b^3+c^3$

$=RHS$

$\therefore LHS=RHS$

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

Since, a, b, c are in G.P.

$\therefore a=a,b=ar,c=a{r}^2$

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

$=\frac{(a+ar+a{r}^2{)}^2}{a^2+a^2{r}^2+a^2{r}^4}$

$=\frac{a^2(1+r+r^2{)}^2}{a^2(1+r^2+r^4)}$

$=\frac{a^2(1+r+r^2{)}^2}{a^2\left[\right(1+r^2-r\left)\right(1+r^2+r\left)\right]}$

$=\frac{a(1+r+r^2)}{a(1+r^2-r)}$

$=\frac{a+ar+a{r}^2}{a+ar-ar}$

$=\frac{a+b+c}{a-b+c}$

= RHS

$\therefore$ LHS = RHS

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

Since, a, b, c are in G.P.

$\therefore$ a = a, b = ar, c = $a{r}^2$

LHS = $\frac{1}{a^2-b^2}+\frac{1}{b^2}$

$=\frac{1}{a^2-a^2{r}^2}+\frac{1}{a^2{r}^2}$

$=\frac{1}{a^2}[\frac{1}{1-r^2}+\frac{1}{r^2}]$

$=\frac{1}{a^2}\left[\frac{r^2+1-r^2}{(1-r^2)r^2}\right]$

$=\frac{1}{a^2}\left[\frac{1}{r^2-r^4}\right]$

$=\frac{1}{(ar{)}^2-(a{r}^2{)}^2}$

$=\frac{1}{b^2-c^2}$

$=RHS$

$\therefore$ LHS = RHS

(v) $(a+2b+2c)(a-2b+2c)=a^2+4c^2$.

Since, a, b, c are in G.P.

$\therefore$ a = a, b = ar, c = $a{r}^2$

LHS = $(a+2b+2c)(a-2b+2c)$

$=(a+2ar+2a{r}^2)(a-2ar+2a{r}^2)$

$=a^2(1+2r+2r^2)(1-2r+2r^2)$

$=a^2(1+2r+2r^2)(1-2r+2r^2)$

$=a^2\left[\right(1+2r^2{)}^2-\left(2r{)}^2\right]$

$=a^2[1+4r^4+4r^2-4r^2]$

$=a^2+4(a{r}^2{)}^2$

$=a^2+4c^2$

= RHS

$\therefore LHS=RHS$