Question

If \[\dfrac{1}{{\sqrt b + \sqrt c }},\dfrac{1}{{\sqrt c + \sqrt a }},\dfrac{1}{{\sqrt a + \sqrt b }}\] are in AP then \[{9^{ax + 1}},{9^{bx + 1}},{9^{cx + 1}}\] \[x \ne 0\]are in
A) GP
B) GP only if \[x < 0\]
C) GP only if \[x > 0\]
D) None of these

Answer 1

Hint: In this question we have to find type series follow by the given terms. Terms are given in AP then use property of AP which state that if three given terms are in series then twice the middle term equal to sum of first and third term.

Formula used: If a,b,c are in AP then
\[2b = a + c\]
If we multiply any constant with all the terms in AP then resultant term will also be in AP
If we add any constant with all terms of AP then resultant term will also be in AP

Complete step by step solution: Given: \[\dfrac{1}{{\sqrt b + \sqrt c }},\dfrac{1}{{\sqrt c + \sqrt a }},\dfrac{1}{{\sqrt a + \sqrt b }}\] are in
We know that If a,b,c are in AP then
\[2b = a + c\]
\[\dfrac{2}{{\sqrt c + \sqrt a }} = \dfrac{1}{{\sqrt b + \sqrt c }} + \dfrac{1}{{\sqrt a + \sqrt b }}\]
\[\dfrac{2}{{\sqrt c + \sqrt a }} = \dfrac{{\sqrt a + 2\sqrt b + \sqrt c }}{{(\sqrt b + \sqrt c )(\sqrt a + \sqrt b )}}\]
\[2\sqrt {ab} + 2b + 2\sqrt {ac} + 2\sqrt {bc} = (\sqrt c + \sqrt a )(\sqrt a + 2\sqrt b + \sqrt c )\]
\[2\sqrt {ab} + 2b + 2\sqrt {ac} + 2\sqrt {bc} = {(\sqrt c + \sqrt a )^2} + 2\sqrt b (\sqrt a + \sqrt c )\]
\[2\sqrt {ab} + 2b + 2\sqrt {ac} + 2\sqrt {bc} = c + a + 2\sqrt {ac} + 2\sqrt {ab} + 2\sqrt {bc} \]
\[2b = a + c\]
a, b, c are also in AP
We know that if we multiply any constant with all the terms in AP then resultant term will also be in AP
Multiply x in each term then resultant term ax, bx, cx will also be in AP
We know that if we add any constant with all terms of AP then resultant term will also be in AP
Add 1 in each ax, bx, cx we get \[ax + 1,bx + 1,cx + 1\]
\[2(bx + 1) = (ax + 1) + (cx + 1)\]………………………….. (i)
Now we have
\[{9^{ax + 1}},{9^{bx + 1}},{9^{cx + 1}}\]
In order to prove the above terms in GP we apply the concept that the square of the middle term is a product of the first and second term.
\[{({9^{bx + 1}})^2} = {9^{2(bx + 1)}}\]

From equation (i)
\[{({9^{bx + 1}})^2} = {9^{(ax + 1) + (cx + 1)}}\]
On simplification we get
\[{({9^{bx + 1}})^2} = {9^{(ax + 1)}} \times {9^{(cx + 1}}\]
The square of the middle term is a product of the first and second term. So we can say that
\[{9^{ax + 1}},{9^{bx + 1}},{9^{cx + 1}}\]are in GP

Thus, Option (A) is correct.

Note: Here we must remember that if we multiply any constant with all the terms in AP then resultant term will also be in AP, if we add any constant with all terms of AP then resultant term will also be in AP. If square of middle term is a product of first term and third term then three terms are in GP.
Sometime students get confused in between AP and GP the only difference in between them is in AP we talk about common difference whereas in GP we talk about common ratio.