If a, b, and c are in arithmetic progression, | vedclass

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

Question

(D) Given that $a, b,$ and $c$ are in arithmetic progression $(AP)$.Therefore,$2b = a + c$.Consider the terms $2^{ax + 1}, 2^{bx + 1},$ and $2^{cx + 1

Vedclass · Accepted Answer

in geometric progression for every $x 
eq 0$.

Vedclass · Answer

(D) Given that $a, b,$ and $c$ are in arithmetic progression $(AP)$.Therefore,$2b = a + c$.Consider the terms $2^{ax + 1}, 2^{bx + 1},$ and $2^{cx + 1}$.For these terms to be in geometric progression $(GP)$,the ratio of consecutive terms must be equal:$\frac{2^{bx + 1}}{2^{ax + 1}} = \frac{2^{cx + 1}}{2^{bx + 1}}$$2^{(b - a)x} = 2^{(c - b)x}$Since $a, b, c$ are in $AP$,we have $b - a = c - b = d$ (common difference).Thus,$2^{dx} = 2^{dx}$,which is always true for any $x 
eq 0$.Therefore,$2^{ax + 1}, 2^{bx + 1},$ and $2^{cx + 1}$ are in geometric progression for every $x 
eq 0$.

If $a, b,$ and $c$ are in arithmetic progression,then $2^{ax + 1}, 2^{bx + 1},$ and $2^{cx + 1}$ for $x \neq 0$ are...

Similar Questions

In which progression are the terms $\frac{1}{1 + \sqrt{x}}, \frac{1}{1 - x}, \frac{1}{1 - \sqrt{x}}$?

Three non-zero real numbers form an $A.P.$ and the squares of these numbers taken in the same order form a $G.P.$ Then the number of all possible common ratios of the $G.P.$ is

If the ratio of $A.M.$ between two positive real numbers $a$ and $b$ to their $H.M.$ is $m:n$,then $a:b$ is

If the ratio of $H.M.$ and $G.M.$ between two numbers $a$ and $b$ is $4:5$,then the ratio of the two numbers will be

If $\frac{a + b}{1 - ab}, b, \frac{b + c}{1 - bc}$ are in $A.P.$,then $a, \frac{1}{b}, c$ are in

Vedclass Test Series

Exam Paper Generator

Online Exam Module