Question

If a, b, c are in G.P. and ${{\text{a}}^{\text{x}}}$=${{\text{b}}^{\text{y}}}$=${{\text{c}}^{\text{z}}}$, prove that $\dfrac{1}{{\text{x}}} + \dfrac{1}{{\text{z}}} = \dfrac{2}{{\text{y}}}$

Answer 1

We know that in G.P. there is a common ratio between the consecutive terms of the series. So we can write b=ar and c=br and also${\text{c = a}}{{\text{r}}^{\text{2}}}$. Now put these values in ${{\text{a}}^{\text{x}}}$=${{\text{b}}^{\text{y}}}$=${{\text{c}}^{\text{z}}}$ and solve. Then take two terms and simplify them to get the values in base a and r on either side. Then take two different terms and solve in the same manner. Compare the results and equate the terms having the same base. Solve it to get the answer.

Complete step-by-step answer:
Given, a, b, c are in G.P. so the terms have a common ratio r which means-
$\Rightarrow \dfrac{{\text{b}}}{{\text{a}}} = {\text{r}}$ and $\dfrac{{\text{c}}}{{\text{b}}} = {\text{r}}$
Then we can write it as-
$\Rightarrow$ b=ar and c=br
But since b=ar so we can write c as-
$\Rightarrow {\text{c = a}}{{\text{r}}^{\text{2}}}$
Also given, ${{\text{a}}^{\text{x}}}$=${{\text{b}}^{\text{y}}}$=${{\text{c}}^{\text{z}}}$ -- (i)
Now, we have to prove that$\dfrac{1}{{\text{x}}} + \dfrac{1}{{\text{z}}} = \dfrac{2}{{\text{y}}}$
So on putting the values of b and c in eq. (i), we get-
$\Rightarrow {{\text{a}}^{\text{x}}} = {\left( {{\text{ar}}} \right)^{\text{y}}} = {\left( {{\text{a}}{{\text{r}}^2}} \right)^{\text{z}}}$
On solving, we get-
$\Rightarrow {{\text{a}}^{\text{x}}} = {{\text{a}}^{\text{y}}}{{\text{r}}^{\text{y}}} = {{\text{a}}^{\text{z}}}{{\text{r}}^2}^{\text{z}}$--- (ii)
On taking the first and second term, we get-
$\Rightarrow {{\text{a}}^{\text{x}}} = {{\text{a}}^{\text{y}}}{{\text{r}}^{\text{y}}}$
On rearrangement, we get-
$\Rightarrow \dfrac{{{{\text{a}}^{\text{x}}}}}{{{{\text{a}}^{\text{y}}}}} = {{\text{r}}^{\text{y}}}$
On solving, we get-
$\Rightarrow {{\text{a}}^{\text{x}}}{{\text{a}}^{{\text{ - y}}}} = {{\text{r}}^{\text{y}}}$
Now, we know if the base of two multiplicative terms is the same, the powers of the base are added and the base remains the same. So we get-
$\Rightarrow {{\text{a}}^{\text{x}}}^{{\text{ - y}}} = {{\text{r}}^{\text{y}}}$-- (iii)
On taking the first and third term, we get-
$\Rightarrow {{\text{a}}^{\text{x}}} = {{\text{a}}^{\text{z}}}{{\text{r}}^2}^{\text{z}}$
On rearrangement, we get-
$\Rightarrow \dfrac{{{{\text{a}}^{\text{x}}}}}{{{{\text{a}}^{\text{z}}}}} = {{\text{r}}^2}^{\text{z}}$
On solving, we get-
$\Rightarrow {{\text{a}}^{\text{x}}}{{\text{a}}^{{\text{ - z}}}} = {{\text{r}}^2}^{\text{z}}$
We know if the base of two multiplicative terms is the same, the powers of the base are added. So we get-
$\Rightarrow {{\text{a}}^{\text{x}}}^{{\text{ - z}}} = {{\text{r}}^2}^{\text{z}}$-- (iv)
Now, since the first, second, and third terms are equal in eq. (ii), then eq. (iii) and (iv) should also be equal to each other. So on comparing the left-hand side part of eq. (iii) and (iv), we get-
$\Rightarrow {{\text{a}}^{\text{x}}}^{{\text{ - y}}} = {{\text{a}}^{\text{x}}}^{{\text{ - z}}}$
Since the base is the same, so the powers will also be equal. So we get-
$\Rightarrow {\text{x - y}} = {\text{x - z}}$-- (v)
Now, on comparing the right-hand side part of eq. (iii) and (iv), we get-
$\Rightarrow {{\text{r}}^{\text{y}}} = {{\text{r}}^2}^{\text{z}}$
Since the base is the same, so the powers will also be equal. So we get-
$\Rightarrow {\text{y = 2z}}$-- (vi)
On dividing eq. (v) by eq. (vi), we get-
$\Rightarrow \dfrac{{{\text{x - y}}}}{{\text{y}}} = \dfrac{{{\text{x - z}}}}{{{\text{2z}}}}$
On cross-multiplication, we get-
$\Rightarrow {\text{2xz - 2yz = xy - yz}}$
On separating common variables, we get-
$\Rightarrow {\text{2xz = xy + 2yz - yz}}$
On solving, we get-
$\Rightarrow {\text{2xz = xy + yz}}$
Now on dividing both sides by xyz, we get-
$\Rightarrow \dfrac{{{\text{2xz}}}}{{{\text{xyz}}}}{\text{ = }}\dfrac{{{\text{xy + yz}}}}{{{\text{xyz}}}}$
On solving, we get-
$\Rightarrow \dfrac{{\text{2}}}{{\text{y}}}{\text{ = }}\dfrac{1}{{\text{z}}} + \dfrac{1}{{\text{x}}}$
Hence proved

Note: Here, we have taken the first and seconds terms and first and third terms in eq. (ii), so that the calculations become easy and we can easily prove the equation given in the question, if we take other terms then the equation we will obtain will become complex and it will take more time to solve.