Sequence and Series
AM-GM Inequality
JEE Main 2020
Grade 11
Question:
The sum of the first three terms of a G.P. is $S$ and their product is 27. Then all possible values of $S$ lie in:
(1) $(-\infty, -9]$
(2) $[-3, \infty)$
(3) $(-\infty, -3] \cup [9, \infty)$
(4) $[9, \infty)$
Step-by-Step Solution
Key Concept: Write the three terms as $a/r, a, ar$. Their product gives $a^3 = 27$ so $a = 3$. Sum $S = 3(1/r + 1 + r) = 3(r + 1/r) + 3$. Use AM-GM: $r + 1/r \geq 2$ for $r > 0$ and $\leq -2$ for $r < 0$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (3)