If the sum of three terms of a G.P. is 19 and | vedclass

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

Question

(B) Let the three terms of the $G.P.$ be $\frac{a}{r}, a, ar$.Given that the product of the terms is $216$:$(\frac{a}{r}) \cdot a \cdot (ar) = 216$$a^

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(B) Let the three terms of the $G.P.$ be $\frac{a}{r}, a, ar$.Given that the product of the terms is $216$:$(\frac{a}{r}) \cdot a \cdot (ar) = 216$$a^3 = 216$$a = 6$Given that the sum of the terms is $19$:$\frac{a}{r} + a + ar = 19$Substitute $a = 6$:$\frac{6}{r} + 6 + 6r = 19$$\frac{6}{r} + 6r = 13$Multiply by $r$:$6 + 6r^2 = 13r$$6r^2 - 13r + 6 = 0$Factorize the quadratic equation:$6r^2 - 9r - 4r + 6 = 0$$3r(2r - 3) - 2(2r - 3) = 0$$(3r - 2)(2r - 3) = 0$Thus,$r = \frac{2}{3}$ or $r = \frac{3}{2}$.

If the sum of three terms of a $G.P.$ is $19$ and their product is $216$,then the common ratio of the series is

Similar Questions

The sum of the series $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} + \dots$ to $9$ terms is

The sum to infinite terms of the series $1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots$ is

If $a, b$ and $c$ are three distinct numbers in $G.P.$ and $a + b + c = xb$,then $x$ cannot be:

If $n$ arithmetic means $a_1, a_2, \dots, a_n$ are inserted between $50$ and $100$ and $n$ harmonic means $h_1, h_2, \dots, h_n$ are inserted between the same two numbers,then $a_2 h_{n-1}$ is equal to

The product $(32)(32)^{1/6}(32)^{1/36} \dots \infty$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module