The first three terms of a geometric progress | vedclass

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

Question

(B) Let the three terms be $\frac{b}{r}, b, br$ where $r$ is the common ratio.Given that the harmonic mean of $a$ and $b$ is $12$:$\frac{2ab}{a+b} = 1

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Let the three terms be $\frac{b}{r}, b, br$ where $r$ is the common ratio.Given that the harmonic mean of $a$ and $b$ is $12$:$\frac{2ab}{a+b} = 12 \implies \frac{2(\frac{b}{r})b}{\frac{b}{r}+b} = 12 \implies \frac{2b}{1+r} = 12 \implies b = 6(1+r) \dots (1)$Given that the harmonic mean of $b$ and $c$ is $36$:$\frac{2bc}{b+c} = 36 \implies \frac{2b(br)}{b+br} = 36 \implies \frac{2br}{1+r} = 36 \implies br = 18(1+r) \dots (2)$Dividing $(2)$ by $(1)$:$\frac{br}{b} = \frac{18(1+r)}{6(1+r)} \implies r = 3$Substituting $r=3$ into $(1)$:$b = 6(1+3) = 24$Since $a = \frac{b}{r} = \frac{24}{3} = 8$.

The first three terms of a geometric progression are $a, b, c$. If the harmonic mean of $a$ and $b$ is $12$ and the harmonic mean of $b$ and $c$ is $36$,then $a = \dots$

Similar Questions

The sum of the series $5.05 + 1.212 + 0.29088 + \dots \infty$ is

Let $x_1, x_2, \ldots, x_6$ be the roots of the polynomial equation $x^6+2 x^5+4 x^4+8 x^3+16 x^2+32 x+64=0$. Then,

If the product of three consecutive terms of a $G.P.$ is $216$ and the sum of their products taken two at a time is $156$,then the numbers are:

If there are $5$ geometric means between $486$ and $\frac{2}{3}$,what is the $4^{th}$ geometric mean?

If $y = x - x^2 + x^3 - x^4 + \dots \infty$,then the value of $x$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module