If the sum and product of four positive conse | vedclass

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

Question

(A) Let the four positive consecutive terms of the $G.P.$ be $\frac{a}{r^3}, \frac{a}{r}, ar, ar^3$ with common ratio $R = r^2$.Given product: $\frac{

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) Let the four positive consecutive terms of the $G.P.$ be $\frac{a}{r^3}, \frac{a}{r}, ar, ar^3$ with common ratio $R = r^2$.Given product: $\frac{a}{r^3} 	imes \frac{a}{r} 	imes ar 	imes ar^3 = a^4 = 1296 \implies a = 6$.Given sum: $\frac{a}{r^3} + \frac{a}{r} + ar + ar^3 = 126$.Substituting $a=6$: $\frac{6}{r^3} + \frac{6}{r} + 6r + 6r^3 = 126 \implies (r^3 + \frac{1}{r^3}) + (r + \frac{1}{r}) = 21$.Let $x = r + \frac{1}{r}$. Then $r^3 + \frac{1}{r^3} = x^3 - 3x$.So,$(x^3 - 3x) + x = 21 \implies x^3 - 2x - 21 = 0$.By inspection,$x=3$ is a root: $27 - 6 - 21 = 0$.Thus,$r + \frac{1}{r} = 3 \implies r^2 - 3r + 1 = 0$.The common ratio is $R = r^2$. From $r^2 - 3r + 1 = 0$,we have $r^2 + 1 = 3r$.Dividing by $r$,$r + \frac{1}{r} = 3$. Squaring gives $r^2 + \frac{1}{r^2} + 2 = 9 \implies r^2 + \frac{1}{r^2} = 7$.The two possible common ratios are the roots of $t^2 - 7t + 1 = 0$ (since $R + \frac{1}{R} = 7$).The sum of the common ratios is $7$.

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 $.........$.

Similar Questions

If the sum of an infinite $GP$ is $9$ and the sum of the first two terms is $5$,then their common ratio is .....

If the roots of the equation $x^5-40x^4-Px^3-Rx-S=0$ are in geometric progression and the sum of the reciprocals of the roots is $10$,then $|S|=$

In a geometric progression,if the ratio of the sum of the first $5$ terms to the sum of their reciprocals is $49$,and the sum of the first and the third term is $35$,then the first term of this geometric progression is:

If in an infinite $G.P.$ the first term is equal to twice the sum of the remaining terms,then its common ratio is

If the roots of the cubic equation $ax^3 + bx^2 + cx + d = 0$ are in $G.P.$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module