If the sum of three terms of a Geometric Prog | vedclass

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

Question

(A) Let the three terms of the $GP$ be $\frac{a}{r}, a, ar$.The product of these terms is $\frac{a}{r} 	imes a 	imes ar = a^3 = 216$.Thus,$a = \sqrt

Vedclass · Accepted Answer

$3/2$

Vedclass · Answer

(A) Let the three terms of the $GP$ be $\frac{a}{r}, a, ar$.The product of these terms is $\frac{a}{r} 	imes a 	imes ar = a^3 = 216$.Thus,$a = \sqrt[3]{216} = 6$.The sum of the three terms is $\frac{6}{r} + 6 + 6r = 19$.Subtracting $6$ from both sides,we get $\frac{6}{r} + 6r = 13$.Multiplying by $r$,we get $6 + 6r^2 = 13r$,which simplifies to the quadratic equation $6r^2 - 13r + 6 = 0$.Factoring the quadratic equation: $6r^2 - 9r - 4r + 6 = 0 \implies 3r(2r - 3) - 2(2r - 3) = 0$.This gives $(3r - 2)(2r - 3) = 0$.Therefore,$r = \frac{2}{3}$ or $r = \frac{3}{2}$.

If the sum of three terms of a Geometric Progression $(GP)$ is $19$ and their product is $216$,then the common ratio of this $GP$ is:

Similar Questions

If $p, q, r$ are in a geometric progression and $a, b, c$ are in another geometric progression,then $cp, bq, ar$ are in...

If the sum and product of four positive consecutive terms of a $G.P.$ are $126$ and $1296$ respectively,then the sum of common ratios of all such $G.P.s$ is $.........$.

The product of three geometric means between $4$ and $\frac{1}{4}$ will be

If every term of a $G.P.$ with positive terms is the sum of its two previous terms,then the common ratio of the series is

Find the $9^{th}$ term of the geometric progression $8 + 12 + 18 + 27 + \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module