JEE MAIN - Mathematics (2020 - 2nd September Morning Slot - No. 7)
The sum of the first three terms of a G.P. is S and
their product is 27. Then all such S lie in :
[-3, $$\infty $$)
(-$$ \propto $$, 9]
(-$$ \propto $$, -9] $$ \cup $$ [-3, $$\infty $$)
(-$$ \propto $$, -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, $$\infty $$)
$$ \therefore $$ $$a\left( {{1 \over r} + 1 + r} \right)$$ = 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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