
Given that \[\alpha ,\gamma \] are the roots of the equation \[A{{x}^{2}}-4x+1=0\] and \[\beta ,\delta \] are the roots of the equation \[B{{x}^{2}}-6x+1=0\], the values of\[A\] and \[B\] such that \[\alpha ,\beta ,\gamma \] and $\delta $ are in H.P. are
A. \[A=3, B=8\]
B. \[A=-3, B=8\]
C. \[A=3, B=-8\]
D. None of these
Answer
163.2k+ views
Hint: In this question, we have to find the values of the variables \[A\] and \[B\]. These values are obtained from the given quadratic expressions by using the given roots \[\alpha,\beta,\gamma \] and $\delta $ which are in H.P. For this, we need to calculate the sum of roots and the product of roots. With these values, we can able to find the required variables.
Formula Used: The relationship between A.P and H.P:
If $a,b,c$ are in A.P then $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ are in H.P.
Or
If $a,b,c$ are H.P then $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ are in H.P.
The general terms in an arithmetic series are $a,a+d,a+2d,...,a+nd$ where the common difference $d={{t}_{n}}-{{t}_{n-1}}$.
If $\alpha ,\beta $ are the roots of a quadratic equation $a{{x}^{2}}+bx+c=0$, then the sum and product of the roots are
$\begin{align}
& \alpha +\beta =\dfrac{-b}{a} \\
& \alpha \cdot \beta =\dfrac{c}{a} \\
\end{align}$
Complete step by step solution: It is given that, \[\alpha ,\beta ,\gamma ,\delta \] are in H.P. So, \[\dfrac{1}{\alpha },\dfrac{1}{\beta },\dfrac{1}{\gamma },\dfrac{1}{\delta }\] are in A.P.
Then, the terms of the series are on comparing with the general terms of A.P, and we get
\[\dfrac{1}{\alpha }=a;\dfrac{1}{\beta }=\dfrac{1}{\alpha }+d;\dfrac{1}{\gamma }=\dfrac{1}{\alpha }+2d;\dfrac{1}{\delta }=\dfrac{1}{\alpha }+3d\]
It is given that, \[\alpha,\gamma \] are the roots of the equation \[A{{x}^{2}}-4x+1=0\].
Then, their sum and products are
\[\begin{align}
& \alpha +\gamma =\dfrac{-(-4)}{A}=\dfrac{4}{A} \\
& \alpha \cdot \gamma =\dfrac{1}{A} \\
\end{align}\]
On dividing the sum and the products, we get
\[\begin{align}
& \dfrac{\alpha +\gamma }{\alpha \cdot \gamma }=\dfrac{{}^{4}/{}_{A}}{{}^{1}/{}_{A}}=4 \\
& \Rightarrow \dfrac{1}{\alpha }+\dfrac{1}{\gamma }=4 \\
\end{align}\]
But we know \[\dfrac{1}{\gamma }=\dfrac{1}{\alpha }+2d\]
Then, on substituting, we get
\[\begin{align}
& \Rightarrow \dfrac{1}{\alpha }+\dfrac{1}{\alpha }+2d=4 \\
& \Rightarrow \dfrac{2}{\alpha }+2d=4 \\
& \Rightarrow \dfrac{1}{\alpha }+d=2\text{ }...(1) \\
\end{align}\]
It is given that, \[\beta ,\delta \] are the roots of the equation \[B{{x}^{2}}-6x+1=0\].
Then, their sum and products are
\[\begin{align}
& \beta +\delta =\dfrac{-(-6)}{B}=\dfrac{6}{B} \\
& \beta \cdot \delta =\dfrac{1}{B} \\
\end{align}\]
On dividing the sum and the products, we get
\[\begin{align}
& \dfrac{\beta +\delta }{\beta \cdot \delta }=\dfrac{{}^{6}/{}_{B}}{{}^{1}/{}_{B}}=6 \\
& \Rightarrow \dfrac{1}{\beta }+\dfrac{1}{\delta }=6 \\
\end{align}\]
But we know \[\dfrac{1}{\beta }=\dfrac{1}{\alpha }+d;\dfrac{1}{\delta }=\dfrac{1}{\alpha }+3d\]
Then, on substituting, we get
\[\begin{align}
& \dfrac{1}{\alpha }+d+\dfrac{1}{\alpha }+3d=6 \\
& \dfrac{2}{\alpha }+4d=6 \\
& \dfrac{1}{\alpha }+2d=3\text{ }...(2) \\
\end{align}\]
Then, from (1) and (2), we get
\[\begin{align}
& \dfrac{1}{\alpha }+d+d=3 \\
& \Rightarrow 2+d=3 \\
& \Rightarrow d=3-2=1 \\
\end{align}\]
Substituting obtained value in (1), we get
\[\begin{align}
& \dfrac{1}{\alpha }+d=2 \\
& \Rightarrow \dfrac{1}{\alpha }+1=2 \\
& \Rightarrow \dfrac{1}{\alpha }=1 \\
\end{align}\]
Thus, the other terms are
\[\begin{align}
& \dfrac{1}{\beta }=\dfrac{1}{\alpha }+d=1+1=2 \\
& \dfrac{1}{\gamma }=\dfrac{1}{\alpha }+2d=1+2(1)=3 \\
& \dfrac{1}{\delta }=\dfrac{1}{\alpha }+3d=1+3(1)=4 \\
\end{align}\]
So, applying these values into the products of the roots of the given equations, we get
\[\begin{align}
& A=\dfrac{1}{\alpha \gamma }=\dfrac{1}{\alpha }\times \dfrac{1}{\gamma }=1\times 3=3 \\
& B=\dfrac{1}{\beta \delta }=\dfrac{1}{\beta }\times \dfrac{1}{\delta }=2\times 4=8 \\
& \therefore A=3,B=8 \\
\end{align}\]
Option ‘A’ is correct
Note: Here we need to remember the rule, if the given terms are in H.P then their reciprocals are in A.P and vice versa. So, by using this concept, we can evaluate the required variables in the above question.
Formula Used: The relationship between A.P and H.P:
If $a,b,c$ are in A.P then $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ are in H.P.
Or
If $a,b,c$ are H.P then $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ are in H.P.
The general terms in an arithmetic series are $a,a+d,a+2d,...,a+nd$ where the common difference $d={{t}_{n}}-{{t}_{n-1}}$.
If $\alpha ,\beta $ are the roots of a quadratic equation $a{{x}^{2}}+bx+c=0$, then the sum and product of the roots are
$\begin{align}
& \alpha +\beta =\dfrac{-b}{a} \\
& \alpha \cdot \beta =\dfrac{c}{a} \\
\end{align}$
Complete step by step solution: It is given that, \[\alpha ,\beta ,\gamma ,\delta \] are in H.P. So, \[\dfrac{1}{\alpha },\dfrac{1}{\beta },\dfrac{1}{\gamma },\dfrac{1}{\delta }\] are in A.P.
Then, the terms of the series are on comparing with the general terms of A.P, and we get
\[\dfrac{1}{\alpha }=a;\dfrac{1}{\beta }=\dfrac{1}{\alpha }+d;\dfrac{1}{\gamma }=\dfrac{1}{\alpha }+2d;\dfrac{1}{\delta }=\dfrac{1}{\alpha }+3d\]
It is given that, \[\alpha,\gamma \] are the roots of the equation \[A{{x}^{2}}-4x+1=0\].
Then, their sum and products are
\[\begin{align}
& \alpha +\gamma =\dfrac{-(-4)}{A}=\dfrac{4}{A} \\
& \alpha \cdot \gamma =\dfrac{1}{A} \\
\end{align}\]
On dividing the sum and the products, we get
\[\begin{align}
& \dfrac{\alpha +\gamma }{\alpha \cdot \gamma }=\dfrac{{}^{4}/{}_{A}}{{}^{1}/{}_{A}}=4 \\
& \Rightarrow \dfrac{1}{\alpha }+\dfrac{1}{\gamma }=4 \\
\end{align}\]
But we know \[\dfrac{1}{\gamma }=\dfrac{1}{\alpha }+2d\]
Then, on substituting, we get
\[\begin{align}
& \Rightarrow \dfrac{1}{\alpha }+\dfrac{1}{\alpha }+2d=4 \\
& \Rightarrow \dfrac{2}{\alpha }+2d=4 \\
& \Rightarrow \dfrac{1}{\alpha }+d=2\text{ }...(1) \\
\end{align}\]
It is given that, \[\beta ,\delta \] are the roots of the equation \[B{{x}^{2}}-6x+1=0\].
Then, their sum and products are
\[\begin{align}
& \beta +\delta =\dfrac{-(-6)}{B}=\dfrac{6}{B} \\
& \beta \cdot \delta =\dfrac{1}{B} \\
\end{align}\]
On dividing the sum and the products, we get
\[\begin{align}
& \dfrac{\beta +\delta }{\beta \cdot \delta }=\dfrac{{}^{6}/{}_{B}}{{}^{1}/{}_{B}}=6 \\
& \Rightarrow \dfrac{1}{\beta }+\dfrac{1}{\delta }=6 \\
\end{align}\]
But we know \[\dfrac{1}{\beta }=\dfrac{1}{\alpha }+d;\dfrac{1}{\delta }=\dfrac{1}{\alpha }+3d\]
Then, on substituting, we get
\[\begin{align}
& \dfrac{1}{\alpha }+d+\dfrac{1}{\alpha }+3d=6 \\
& \dfrac{2}{\alpha }+4d=6 \\
& \dfrac{1}{\alpha }+2d=3\text{ }...(2) \\
\end{align}\]
Then, from (1) and (2), we get
\[\begin{align}
& \dfrac{1}{\alpha }+d+d=3 \\
& \Rightarrow 2+d=3 \\
& \Rightarrow d=3-2=1 \\
\end{align}\]
Substituting obtained value in (1), we get
\[\begin{align}
& \dfrac{1}{\alpha }+d=2 \\
& \Rightarrow \dfrac{1}{\alpha }+1=2 \\
& \Rightarrow \dfrac{1}{\alpha }=1 \\
\end{align}\]
Thus, the other terms are
\[\begin{align}
& \dfrac{1}{\beta }=\dfrac{1}{\alpha }+d=1+1=2 \\
& \dfrac{1}{\gamma }=\dfrac{1}{\alpha }+2d=1+2(1)=3 \\
& \dfrac{1}{\delta }=\dfrac{1}{\alpha }+3d=1+3(1)=4 \\
\end{align}\]
So, applying these values into the products of the roots of the given equations, we get
\[\begin{align}
& A=\dfrac{1}{\alpha \gamma }=\dfrac{1}{\alpha }\times \dfrac{1}{\gamma }=1\times 3=3 \\
& B=\dfrac{1}{\beta \delta }=\dfrac{1}{\beta }\times \dfrac{1}{\delta }=2\times 4=8 \\
& \therefore A=3,B=8 \\
\end{align}\]
Option ‘A’ is correct
Note: Here we need to remember the rule, if the given terms are in H.P then their reciprocals are in A.P and vice versa. So, by using this concept, we can evaluate the required variables in the above question.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main 2025 Session 2: Exam Date, Admit Card, Syllabus, & More

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Trending doubts
Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

JEE Main Chemistry Question Paper with Answer Keys and Solutions

JEE Main Reservation Criteria 2025: SC, ST, EWS, and PwD Candidates

What is Normality in Chemistry?

Chemistry Electronic Configuration of D Block Elements: JEE Main 2025

Other Pages
NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series

Total MBBS Seats in India 2025: Government College Seat Matrix

NEET Total Marks 2025: Important Information and Key Updates

Neet Cut Off 2025 for MBBS in Tamilnadu: AIQ & State Quota Analysis

Karnataka NEET Cut off 2025 - Category Wise Cut Off Marks
