Question

If ${{a}_{1}},{{a}_{2}},{{a}_{3,}}\left( {{a}_{1}}>0 \right)$ are in GP with common ratio $r$, for which the inequality$9{{a}_{1}}+5{{a}_{3}}>14{{a}_{2}}$ holds, cannot lie in the interval
1.$\left[ 1,\infty \right]$
2.$\left[ 1,\dfrac{9}{5} \right]$
3.$\left[ \dfrac{4}{5},1 \right]$
4.$\left[ \dfrac{5}{9},1 \right]$

Answer 1

In order to solve it, we will be assuming both the first terms and the common ratio. Then we will be considering the given inequality by transposing the term from RHS to LHS. We will be expressing the second and the third terms in the product of the first term and common ratio. And upon solving it, we will be finding for which interval it does not hold.

Complete step-by-step solution:
Now let us learn more about the geometric progression. It is a type of sequence in which each succeeding term is produced by multiplying the preceding term with a fixed number which is termed as the common ratio. The general form of the sequence is $a,ar,a{{r}^{2}},a{{r}^{3}}...$ where $a$ is the first term and $r$ is the common ratio. The formula for finding the ${{n}^{th}}$ term of a GP is ${{a}^{n}}=a{{r}^{n-1}}$. We can also find the sum of the sequence which is in GP.
Now let us solve the given problem.
We are given with the inequality i.e. $9{{a}_{1}}+5{{a}_{3}}>14{{a}_{2}}$.
Firstly, let us consider that first term as ${{a}_{1}}$ and the common ratio would be $r$.
Now we can determine the second and the third term and they are-

& {{a}_{2}}={{a}_{1}}r \\\ & {{a}_{3}}={{a}_{1}}{{r}^{2}} \\\ \end{aligned}$$ So we can express it in the following way. $$5{{a}_{3}}=5{{a}_{1}}{{r}^{2}}$$ $$14{{a}_{2}}=14{{a}_{1}}r$$ Now let us substitute them, we get $$\begin{aligned} & 9{{a}_{1}}+5{{a}_{3}}>14{{a}_{2}} \\\ & \Rightarrow 9{{a}_{1}}+5{{a}_{1}}{{r}^{2}}>14{{a}_{1}}r \\\ \end{aligned}$$ Upon taking $${{a}_{1}}$$ common and transposing, we get $$\begin{aligned} & 9{{a}_{1}}+5{{a}_{3}}>14{{a}_{2}} \\\ & \Rightarrow 9{{a}_{1}}+5{{a}_{1}}{{r}^{2}}>14{{a}_{1}}r \\\ & \Rightarrow 5{{r}^{2}}-14r+9>0 \\\ \end{aligned}$$ Now let us solve for quadratic inequality. We get $$\begin{aligned} & \Rightarrow 5{{r}^{2}}-14r+9>0 \\\ & \Rightarrow 5{{r}^{2}}-5r-9r+9>0 \\\ & \Rightarrow 5r\left( r-1 \right)-9\left( r-1 \right)>0 \\\ & \Rightarrow \left( r-1 \right)\left( 5r-9 \right)>0 \\\ \end{aligned}$$ Now let us find the interval of $$r$$. We get $$r\in \left( -\infty ,1 \right)\cup \left( \dfrac{9}{5},\infty \right)$$ So we can conclude that $$r$$ does not lie in the interval $$\left[ 1,\dfrac{9}{5} \right]$$. **Hence option 2 is the correct answer.** **Note:** We must know that the sequence that follows the geometric progression is called a geometric sequence. While finding the interval if it holds or not, we must solve only the equation or the inequality given. We must express other terms in the product of the first term and common ratio for our easy calculation. This concept can be applied to one of the famous examples i.e. filling the chess board with rice grains.