$(x + 1), (x + 5), (x + 7)$ and $(x + 19)$ are in proportion find value of $x$ .

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Solution

The correct option is A $2$
In, $(x + 1) : (x + 5) :: (x + 7) : (x + 19)$ , product of extremes $=$ product of means
$(x + 1) \times (x + 19) = (x + 5) \times (x + 7)$
$x^{2} + x + 19 x + 19 = x^{2} + 5 x + 7 x + 35$
$20 x + 19 = 12 x + 35$
$8 x = 16$
$x = 2$

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