If alpha, beta, gamma are the geometric means | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given that $\alpha, \beta, \gamma$ are geometric means between $ca, ab$; $ab, bc$; and $bc, ca$ respectively.Therefore,$\alpha^2 = (ca)(ab) = a^2b

Vedclass · Accepted Answer

$A.P.$

Vedclass · Answer

(A) Given that $\alpha, \beta, \gamma$ are geometric means between $ca, ab$; $ab, bc$; and $bc, ca$ respectively.Therefore,$\alpha^2 = (ca)(ab) = a^2bc$,$\beta^2 = (ab)(bc) = b^2ca$,and $\gamma^2 = (bc)(ca) = c^2ab$.Since $a, b, c$ are in $A.P.$,we have $2b = a + c$.We need to check if $\alpha^2, \beta^2, \gamma^2$ are in $A.P.$,which requires $2\beta^2 = \alpha^2 + \gamma^2$.Substituting the values: $2(b^2ca) = a^2bc + c^2ab$.Dividing both sides by $abc$ (assuming $a, b, c 
eq 0$),we get $2b^2 = ac + ac$ is incorrect; rather,$2b^2ca = abc(a + c)$.Dividing by $abc$,we get $2b = a + c$,which is true as $a, b, c$ are in $A.P$.Thus,$\alpha^2, \beta^2, \gamma^2$ are in $A.P.$

If $\alpha, \beta, \gamma$ are the geometric means between $ca, ab$; $ab, bc$; and $bc, ca$ respectively,where $a, b, c$ are in $A.P.$,then $\alpha^2, \beta^2, \gamma^2$ are in

Similar Questions

If the $12^{th}$ term of an $A.P.$ is $-13$ and the sum of the first four terms is $24,$ then what is the sum of the first $10$ terms?

The sum of the first $20$ terms common between the series $3 + 7 + 11 + 15 + \dots$ and $1 + 6 + 11 + 16 + \dots$ is:

Let $a_{1}, a_{2}, a_{3}, \ldots$ be a $G$.$P$. such that $a_{1} < 0$; $a_{1} + a_{2} = 4$ and $a_{3} + a_{4} = 16$. If $\sum_{i=1}^{9} a_{i} = 4 \lambda$,then $\lambda$ is equal to:

Let $a_1, a_2, ..., a_{10}$ be a $G.P.$ If $\frac{a_3}{a_1} = 25$,then $\frac{a_9}{a_5}$ is equal to:

The value of $0.2overline{34}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module