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

[1, \infty)

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[1, 9 / 5]

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[4 / 5, 1]

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[5 / 9, 1]

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Solution

The correct option is A $[1, 9 / 5]$
$a_{1}, a_{2}, a_{3} - G . P$
common ration = r
$∴ a_{2} = a_{1} r a_{3} = a_{1} r^{2} 9 a_{1} + 5 a_{3} > 14 a_{2} 9 a_{1} - 14 a_{1} r + 9 > 0 5 r^{2} - 9 r - 5 r + 9 > 0 r (5 r - 9) - 1 (5 r - 9) > 0 (r - 1) (5 r - 9) > 0 (r - 1) (r - \frac{9}{5}) > 0 r \in (- \infty, 1) \cup (\frac{9}{5}, \infty) ∴ r \notin [1, \frac{9}{5}]$

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

The correct option is A [1,9/5]a1,a2,a3−G.Pcommon ration = r∴a2=a1ra3=a1r29a1+5a3>14a29a1−14a1r+9>05r2−9r−5r+9>0r(5r−9)−1(5r−9)>0(r−1)(5r−9)>0(r−1)(r−95)>0r∈(−∞,1)∪(95,∞)∴r∉[1,95]

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