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Maths-

General

Easy

Question

If $a subscript 1 end subscript comma a subscript 2 end subscript comma a subscript 3 end subscript left parenthesis a subscript 1 end subscript greater than 0 right parenthesis$ are three successive terms of a G.P. with common ratio $r$ , the value of $r$ for which $a subscript 3 end subscript greater than 4 a subscript 2 end subscript minus 3 a subscript 1 end subscript$ holds is given by

$1 < r < 3$
$- 3 < r < - 1$
$r > 3$ or $r < 1$
None of these

The correct answer is: $r greater than 3$ or $r less than 1$

Let $a subscript 1 end subscript comma blank a subscript 2 end subscript$ , and $a subscript 3 end subscript$ be first three consecutive terms of G.P. with common ratio $r$ . Then,
$a subscript 2 end subscript equals a subscript 1 end subscript r$ and $a subscript 3 end subscript equals a subscript 1 end subscript r to the power of 2 end exponent$
Now, $a subscript 3 end subscript greater than 4 a subscript 2 end subscript minus 3 a subscript 1 end subscript$
$rightwards double arrow a subscript 1 end subscript r to the power of 2 end exponent greater than 4 a subscript 1 end subscript r minus 3 a subscript 1 end subscript$
$rightwards double arrow r to the power of 2 end exponent greater than 4 r minus 3$
$rightwards double arrow r to the power of 2 end exponent minus 4 r plus 3 greater than 0$
$rightwards double arrow open parentheses r minus 1 close parentheses open parentheses r minus 3 close parentheses greater than 0$
$rightwards double arrow r less than 1$ or $r greater than 3$

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