Question

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

(i) $(a^2+b^2),(b^2+c^2),(c^2+d{)}^2$ are in G.P.

(ii) $(a^2-b^2),(b^2-c^2),(c^2-d{)}^2$ are in G.P.

(iii) $\frac{1}{a^2+b^2},\frac{1}{b^2+c^2},\frac{1}{c^2+d^2}$ are in G.P.

(iv) $(a^2+b^2+c^2),(ab+bc+cd),(b^2+c^2+d^2)$

Answer 1

(i) a, b, c, d are in G.P.

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

Now,

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

$(a^2{r}^2+a^2{r}^4{)}^2=(a^2+a^2{r}^2\left)\right(a^2{r}^4+a^2{r}^6)$

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

$a^4{a}^4(1+r^2{)}^2=(a^2+b^2\left)\right(c^2+d^2)$

$\Rightarrow (a^2+b^2),(b^2+c^2),(c^2+d^2)$ are in G.P.

(ii) a, b, c, d are in G.P.

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

Now,

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

$(a^2{r}^2-a^2{r}^4{)}^2=(a^2-a^2{r}^2\left)\right(a^2{r}^4-a^2{r}^6)$

$a^4(r^2-r^4{)}^2=a^2(1-r^2\left)a^2{r}^2\right(1-r^2)$

$a^4{a}^4(1-r^2{)}^2=a^4{r}^4(1-r^2{)}^2$

LHS = RHS

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

$\Rightarrow (a^2-b^2),(b^2-c^2),(c^2-d^2)$ are in G.P.

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

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

Now,

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

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

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

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

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

LHS = RHS

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

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

(iv) a, b, c, d are in G.P.

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

Now,

$(ab+bc+cd{)}^2=(a^2+b^2+c^2\left)\right(b^2+c^2+d^2)$

$(a^{2}r+a^2{r}^3+a^2{r}^5)=(a^2+a^2{r}^2+a^2{r}^4)(a^2+a^2{r}^4+a^2{r}^6)$

$a^4(r+r^3+r^5{)}^2=a^2(1+r^2+r^4\left)a^2{r}^4\right(1+r^2+r^4)$

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

LHS = RHS

$\Rightarrow (ab+bc+cd{)}^2=(a^2+b^2+c^2\left)\right(b^2+c^2+d^2)$

$\Rightarrow (a^2+b^2+c^2),(ab+bc+cd),(b^2+c^2+d^2)$ are in G.P.