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The sum of the first three terms of a G.P. is S and their product is 27. Then all such S lie in

(1) $(–\infty,\ –3] \cup [9,\ \infty)$

(2) $(–\infty,\ –9] \cup [3,\ \infty)$

(3) $[–3,\ \infty)$

(4) $(–\infty,\ 9]$

2 Answers

Best answer

Ans. (1) $(-\infty,-3]\cup[9,\infty)$

Sol. Let terms are $a\over r$, a, ar

then a3 = 27 $\implies$ a = 3

Now $3\over r$ + 3 + 3r = S

$3\left({1\over r}+r\right)+3=S$

$r+{1\over r}\geq2$

$3\left({1\over r}+r\right)+3\in(-\infty,-3]\cup[9,\infty)$

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Why r+1/r become> 2
option A is the answer for this question asked in jee mains 2020 september shift 2
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Are you sure Shift 2?
Should I change the tag to shift 2 or the tag is right?

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