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Maths-

General

Easy

Question

If $x comma blank y comma$ and $z$ are in G.P., and $x plus 3 comma blank y plus 3$ , and $z plus 3$ are in H.P., then

$y = 2$
$y = 3$
$y = 1$
$y = 0$

The correct answer is: $y equals 3$

$x comma blank y comma blank z$ are in G.P. Hence,
$y equals x z$
Now, $x plus 3 comma blank y plus 3 comma blank z plus 3$ are in H.P. Hence,
$y plus 3 equals fraction numerator 2 open parentheses x plus 3 close parentheses left parenthesis z plus 3 right parenthesis over denominator open parentheses x plus 3 close parentheses plus left parenthesis z plus 3 right parenthesis end fraction$
$equals fraction numerator 2 left square bracket x z plus 3 open parentheses x plus z close parentheses plus 9 right square bracket over denominator left square bracket open parentheses x plus z close parentheses plus 6 right square bracket end fraction$
$equals fraction numerator 2 open square brackets y to the power of 2 end exponent plus 3 open parentheses x plus z close parentheses plus 9 close square brackets over denominator open square brackets x plus z plus 6 close square brackets end fraction$
Obviously, $y equals 3$ satisfies it

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