If a is the arithmetic mean of b and c,and G_ | vedclass

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

Question

(B) Given that $a$ is the arithmetic mean of $b$ and $c$,we have $a = \frac{b+c}{2}$,so $b+c = 2a$.Since $G_1, G_2$ are two geometric means between $b

Vedclass · Accepted Answer

$2 G_1 G_2 a$

Vedclass · Answer

(B) Given that $a$ is the arithmetic mean of $b$ and $c$,we have $a = \frac{b+c}{2}$,so $b+c = 2a$.Since $G_1, G_2$ are two geometric means between $b$ and $c$,the sequence $b, G_1, G_2, c$ is in Geometric Progression $(GP)$.Let the common ratio be $r$. Then $G_1 = br$,$G_2 = br^2$,and $c = br^3$.Thus,$r = (c/b)^{1/3}$.$G_1 = b(c/b)^{1/3} = b^{2/3} c^{1/3}$ and $G_2 = b(c/b)^{2/3} = b^{1/3} c^{2/3}$.Now,$G_1^3 + G_2^3 = (b^{2/3} c^{1/3})^3 + (b^{1/3} c^{2/3})^3 = b^2 c + b c^2 = bc(b+c)$.Substituting $b+c = 2a$,we get $G_1^3 + G_2^3 = bc(2a)$.Since $G_1 G_2 = (b^{2/3} c^{1/3})(b^{1/3} c^{2/3}) = bc$,we have $G_1^3 + G_2^3 = 2 a G_1 G_2$.

If $a$ is the arithmetic mean of $b$ and $c$,and $G_1, G_2$ are the two geometric means between them,then $G_1^3 + G_2^3 = $

Similar Questions

The sums of $n$ terms of two arithmetic series are in the ratio $(2n + 3) : (6n + 5)$. Find the ratio of their $13^{th}$ terms.

The sum of the series $1^2 \cdot 2 + 2^2 \cdot 3 + 3^2 \cdot 4 + \dots$ to $n$ terms is:

The sum of a few terms of a geometric progression is $728$. If the common ratio is $3$ and the last term is $486$,then the first term of the series will be:

If $\frac{a^{n + 1} + b^{n + 1}}{a^n + b^n}$ is the harmonic mean between $a$ and $b$,then the value of $n$ is

Jairam purchased a house for Rs. $15000$ and paid Rs. $5000$ at once. He promised to pay the remaining amount in annual installments of Rs. $1000$ with $10\%$ per annum interest on the outstanding balance. What is the total amount paid by Jairam in Rs.?

Vedclass Test Series

Exam Paper Generator

Online Exam Module