The sum of the first three terms of a G.P. is S and their product is 27. Then all such S lie in :
Solution
Let three terms of G.P. are ${a \over r}$, a, ar
<br><br>$\therefore$ $a\left( {{1 \over r} + 1 + r} \right)$ = S ...(1)
<br><br>and a<sup>3</sup>
= 27
<br><br>$\Rightarrow$ a = 3
<br><br>$\therefore$ $3\left( {{1 \over r} + 1 + r} \right)$ = S
<br><br>$\Rightarrow$ ${{1 \over r} + r = {S \over 3} - 1}$
<br><br>$\Rightarrow$ As ${{1 \over r} + r \ge 2}$ or ${{1 \over r} + r \le - 2}$
<br><br>$\therefore$ ${{S \over 3} - 1 \ge 2}$ or ${{S \over 3} - 1 \le - 2}$
<br><br>$\Rightarrow$ ${{S \over 3} \ge 3}$ or ${{S \over 3} \le - 1}$
<br><br>$\Rightarrow$ S $\ge$ 9 or S$\le$ -3
<br><br>$\therefore$ S $\in$ (-$\propto$, -3] $\cup$ [9, $\infty$)
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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