If the product of three terms of a G.P. is 51 | vedclass

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

Question

(B) Let the three terms of a $G.P.$ be $\frac{a}{r}, a, ar$.Given that the product is $512$:$\frac{a}{r} 	imes a 	imes ar = 512$$a^3 = 512 = 8^3$$a

Vedclass · Accepted Answer

$4, 8, 16$

Vedclass · Answer

(B) Let the three terms of a $G.P.$ be $\frac{a}{r}, a, ar$.Given that the product is $512$:$\frac{a}{r} 	imes a 	imes ar = 512$$a^3 = 512 = 8^3$$a = 8$.According to the second condition,$\frac{a}{r} + 8, a + 6, ar$ are in $A.P.$Substituting $a = 8$:$\frac{8}{r} + 8, 8 + 6, 8r$ are in $A.P.$$\frac{8}{r} + 8, 14, 8r$ are in $A.P.$For three terms $x, y, z$ to be in $A.P.$,$2y = x + z$:$2(14) = (\frac{8}{r} + 8) + 8r$$28 = \frac{8}{r} + 8 + 8r$$20 = \frac{8}{r} + 8r$Divide by $4$:$5 = \frac{2}{r} + 2r$$2r^2 - 5r + 2 = 0$$(2r - 1)(r - 2) = 0$$r = 2$ or $r = \frac{1}{2}$.If $r = 2$,the terms are $\frac{8}{2}, 8, 8(2) = 4, 8, 16$.If $r = \frac{1}{2}$,the terms are $\frac{8}{1/2}, 8, 8(1/2) = 16, 8, 4$.Both sets satisfy the condition. Thus,the numbers are $4, 8, 16$ or $16, 8, 4$.

If the product of three terms of a $G.P.$ is $512$. If $8$ is added to the first term and $6$ is added to the second term,the resulting numbers are in $A.P.$. Find the numbers.

Similar Questions

The third term of a $G.P.$ is the square of the first term. If the second term is $8$,then the $6^{th}$ term is:

If $1+\cos x+\cos ^2 x+\cos ^3 x+\ldots \text{ to } \infty = 4+2 \sqrt{3}$,then $x=$

Given a $G.P.$ with $a=729$ and $7^{\text{th}}$ term $64,$ determine $S_{7}$.

If there are $n$ geometric means between $a$ and $b$,what is the common ratio?

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