If $a_{1}, a_{2}, a_{3} (a_{1} > 0)$ are in GP with common ratio $r$ , then the value of $r$ , for which the inequality $9 a_{1} + 5 a_{3} > 14 a_{2}$ holds, cannot lie in the interval

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Solution

Finding the value of $r$ :
Let $a_{1}$ be the first term, $r$ be the common ratio
$a_{2} = a_{1} r a_{3} = a_{1} r^{2}$
Given that
$9 a_{1} + 5 a_{3} > 14 a_{2} \Rightarrow 9 a_{1} + 5 a_{1} r^{2} > 14 a_{1} r \Rightarrow 9 - 14 r + 5 r^{2} > 0 \Rightarrow 5 r^{2} - 5 r - 9 r + 9 > 0 \Rightarrow (r - 1) (5 r - 9) > 0 r \in (- \infty, 1) \cup (\frac{9}{5}, \infty)$
Hence, the value of $r$ does not lie in $[1, \frac{9}{5}] .$

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If a1,a2,a3(a1>0) are in GP with common ratio r, then the value of r, for which the inequality 9a1+5a3>14a2 holds, cannot lie in the interval

Finding the value of r :Let a1 be the first term,r be the common ratioa2=a1ra3=a1r2Given that 9a1+5a3>14a2⇒9a1+5a1r2>14a1r⇒9-14r+5r2>0⇒5r2-5r-9r+9>0⇒(r-1)(5r-9)>0r∈(-∞,1)∪(95,∞)Hence, the value of r does not lie in [1,95].

If $a_{1}, a_{2}, a_{3} (a_{1} > 0)$ are in GP with common ratio $r$ , then the value of $r$ , for which the inequality $9 a_{1} + 5 a_{3} > 14 a_{2}$ holds, cannot lie in the interval