Question

Find the value of $r$ for which the inequality $9{a_1} + 5{a_3} > 14{a_2}$ holds, cannot lie in which of the following interval, given that ${a_1}$ , ${a_2}$ , ${a_3}$ $\left( {{a_1} > 0} \right)$ are in Geometric progression and $r$ is the common ratio.
A. $\left[ {1,\infty } \right]$
B. $\left[ {1,\dfrac{9}{5}} \right]$
C. $\left[ {\dfrac{4}{5},1} \right]$
D. $\left[ {\dfrac{5}{9},1} \right]$

Answer 1

Hint: Write ${a_2}$ , ${a_3}$ in terms of ${a_1}$ and common ratio $r$ and then put the values in the given inequality and then solve it as an algebraic equation.

Formula Used:
General formula of geometric progression is ${t_n} = {t_1} \times {r^{n - 1}}$ , where ${t_1}$ is the first term and $r$ is the common ratio.

Complete step by step solution:
Let ${a_1}$ be the first term, $r$ be the common ratio:
Second term ${a_2} = {a_1}r$
Third term ${a_3} = {a_1}{r^2}$
Given, we have: $9{a_1} + 5{a_3} > 14{a_2}$
Now, substitute the values of ${a_2}$ , ${a_3}$ we have:
$ \Rightarrow 9{a_1} + 5{a_1}{r^2} > 14{a_1}{r_{}}$
$ \Rightarrow 9 - 14r + 5{r^2} > 0$
$ \Rightarrow 5{r^2} - 9r - 5r + 9 > 0$
$ \Rightarrow r(5r - 9) - (5r - 9) > 0$
$ \Rightarrow (r - 1)(5r - 9) > 0$
Therefore, $r \in \left( { - \infty ,1} \right) \cup \left( {\dfrac{9}{5},\infty } \right)$
So, $r$ does not lie in $\left[ {1,\dfrac{9}{5}} \right]$

Option ‘B’ is correct

Note: The grouping is mathematical and results from the quantity of mathematical movement (G.P.). A mathematical series is the sum of a large number of terms in a mathematical configuration. Know a GP well before attempting to determine how to observe the amount of a specific geometric progression. A grouping is referred to as a mathematical movement if, in a series of terms, each subsequent term is created by replicating each preceding word with a constant value. (GP), even though the regular proportion is also known as the consistent worth.