Question

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

(i) $\frac{ab-cd}{b^2-c^2}=\frac{a+c}{b}$

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

(iii) $(b+c)(b+d)=(c+d)(c+d)$

Answer 1

(i) $\frac{ab-cd}{b^2-c^2}=\frac{a+c}{b}$

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

$\therefore a=a,b=ar,c=a{r}^2,d=a{r}^3$

Given, $\frac{ab-cd}{b^2-c^2}=\frac{a+c}{b}$

$\frac{a\left(ar\right)-\left(a{r}^2\right)\left(a{r}^3\right)}{a^2{r}^2-a^2{r}^4}=\frac{a+a{r}^2}{ar}$

$\frac{a^{2}r-a^2{r}^5}{a^2{r}^2(1-r^2)}=\frac{a(1+r^2)}{ar}$

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

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

$\to \frac{1+r^2}{r}=\frac{1+r^2}{r}$

LHS = RHS

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

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

a = a, b = ar, c = $a{r}^2$, d = $a{r}^3$

$\therefore$ Given $(a+b+c+d{)}^2=(a+b{)}^2+2(b+c{)}^2+(c+d{)}^2$

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

$\Rightarrow a^2(1+r+r^2+r^3{)}^2=a^2[(1+r{)}^2+2(r+r^2{)}^2+(r^2+r^3)2]$

$\Rightarrow (1+r+r^2+r^3{)}^2=1+r^2+2r+2(r^2+r^4+2r^3)+r^4+r^6+2r^5$

$\Rightarrow (1+r+r^2+r^3+r+r^2+r^3+r^4+r^2+r^3+r^4+r^5+r^3+r^4+r^5+r^6)$

$=1+r^2+2r+2r^2+2r^4+4r^3+r^4+r^6+2r^5$

$=\Rightarrow (r^6+2r^5+3r^4+4r^3+3r^2+2r+1)$

$=r^6+2r^5+3r^4+4r^3+3r^2+2r+1$

LHS = RHS

(iii) Since, a, b, c are in G.P.

$\therefore$ a = a, b = ar, c = $a{r}^2$, d = $a{r}^3$

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

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

$\Rightarrow a^2(r+r^2)(r+r^3)=a^2(r^2+1)(r^2+r^3)$

$\Rightarrow r^2(1+r)(1+r^2)=r^2(1+r^2)(1+r)$

$\therefore$ LHS = RHS