Let a, b, c 1. If a^3, b^3, c^3 are in A.P. | vedclass

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

Question

(B) Given $a^3, b^3, c^3$ are in $A.P.$,we have $a^3 + c^3 = 2b^3$ $(1)$.Given $\log_a b, \log_c a, \log_b c$ are in $G.P.$,we have $(\log_c a)^2 = (\

Vedclass · Accepted Answer

$216$

Vedclass · Answer

(B) Given $a^3, b^3, c^3$ are in $A.P.$,we have $a^3 + c^3 = 2b^3$ $(1)$.Given $\log_a b, \log_c a, \log_b c$ are in $G.P.$,we have $(\log_c a)^2 = (\log_a b)(\log_b c)$.Using the change of base formula,$(\frac{\ln a}{\ln c})^2 = (\frac{\ln b}{\ln a})(\frac{\ln c}{\ln b}) = \frac{\ln c}{\ln a}$.Thus,$(\ln a)^3 = (\ln c)^3$,which implies $a = c$.Substituting $a = c$ into $(1)$,we get $2a^3 = 2b^3$,so $a = b = c$.The first term of the $A.P.$ is $T_1 = \frac{a+4a+a}{3} = 2a$.The common difference is $d = \frac{a-8a+a}{10} = \frac{-6a}{10} = -\frac{3}{5}a$.The sum of the first $20$ terms is $S_{20} = \frac{20}{2} [2(2a) + (20-1)(-\frac{3}{5}a)] = -444$.$10 [4a - \frac{57}{5}a] = -444$.$10 [\frac{20a - 57a}{5}] = -444$.$2(-37a) = -444$ $\Rightarrow -74a = -444$ $\Rightarrow a = 6$.Since $a = b = c = 6$,then $abc = 6^3 = 216$.

Let $a, b, c > 1$. If $a^3, b^3, c^3$ are in $A.P.$ and $\log_a b, \log_c a, \log_b c$ are in $G.P.$,and the sum of the first $20$ terms of an $A.P.$ with first term $\frac{a+4b+c}{3}$ and common difference $\frac{a-8b+c}{10}$ is $-444$,then $abc$ is equal to:

Similar Questions

Let $S_{n}(x) = \log_{a^{1/2}} x + \log_{a^{1/3}} x + \log_{a^{1/6}} x + \log_{a^{1/11}} x + \log_{a^{1/18}} x + \log_{a^{1/27}} x + \ldots$ up to $n$-terms,where $a > 1$. If $S_{24}(x) = 1093$ and $S_{12}(2x) = 265$,then the value of $a$ is equal to ..... .

Five numbers are in an $AP$ with a common difference $d \neq 0$. If the $1^{st}$, $3^{rd}$, and $4^{th}$ terms are in a $GP$, then:

Suppose the sequence $a_1, a_2, a_3, \ldots$ is an arithmetic progression of distinct numbers such that the sequence $a_1, a_2, a_4, a_8, \ldots$ is a geometric progression. The common ratio of the geometric progression is

If $a_1 = a_2 = 2$ and $a_n = a_{n-1} - 1$ for $n > 2$,then $a_5$ is

Let $3, 7, 11, 15, \ldots, 403$ and $2, 5, 8, 11, \ldots, 404$ be two arithmetic progressions. Then the sum of the common terms in them is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module