
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
162.9k+ views
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.
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.
Recently Updated Pages
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More
