Find the mean of $x$ , $x + 2$ , $x + 4$ , $x + 6$ and $x + 8$

x + 5

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x + 3

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x + 1

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None of these

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Solution

The correct option is B None of these
Given the series: $x, x + 2, x + 4, x + 6, x + 8$
Mean = $\frac{S u m}{C o u n t o f n u m b e r s}$
Mean = $\frac{x + x + 2 + x + 4 + x + 6 + x + 8}{5}$
Mean = $\frac{5 x + 20}{5}$ = $x + 4$

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$6 x^{2} + 17 x + 5$ =?

(a) $(2 x + 1)$

(b) $(2 x + 5) (3 x + 1)$

(d) None of these

The factors of $4 x^{3} + 12 x^{2} + 6 x + 18$ are:

5

x, x + 2, x + 4, x + 6, x + 8 = 13

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