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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 :

JEE · Math · previous-year question

  1. A.[-3, $$\infty $$)
  2. B.(-$$ \propto $$, 9]
  3. C.(-$$ \propto $$, -9] $$ \cup $$ [-3, $$\infty $$)
  4. D.(-$$ \propto $$, -3] $$ \cup $$ [9, $$\infty $$)correct

Answer

D. (-$$ \propto $$, -3] $$ \cup $$ [9, $$\infty $$)

Explanation

Let three terms of G.P. are $${a \over r}$$, a, ar $$ \therefore $$ $$a\left( {{1 \over r} + 1 + r} \right)$$ = S ...(1) and a3 = 27 $$ \Rightarrow $$ a = 3 $$ \therefore $$ $$3\left( {{1 \over r} + 1 + r} \right)$$ = S $$ \Rightarrow $$ $${{1 \over r} + r = {S \over 3} - 1}$$ $$ \Rightarrow $$ As $${{1 \over r} + r \ge 2}$$ or $${{1 \over r} + r \le - 2}$$ $$ \therefore $$ $${{S \over 3} - 1 \ge 2}$$ or $${{S \over 3} - 1 \le - 2}$$ $$ \Rightarrow $$ $${{S \over 3} \ge 3}$$ or $${{S \over 3} \le - 1}$$ $$ \Rightarrow $$ S $$ \ge $$ 9 or S$$ \le $$ -3 $$ \therefore $$ S $$ \in $$ (-$$ \propto $$, -3] $$ \cup $$ [9, $$\infty $$)

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