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Arithmetic Mean

If S A =2 m +...

Question

If $S_{A} = 2 m + (2 m + 1) + (2 m + 2) + . . . + 4 m$ and $S_{B} = (2 m + 1) + (2 m + 3) + (2 m + 5) + . . . + (4 m - 1),$ then

A

\frac{S_{A}}{S_{B}} = 2 + \frac{1}{m}

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B

\frac{S_{A}}{S_{B}} = \frac{4}{3} + \frac{2}{3 m}

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C

\frac{S_{A}}{S_{B}} = \frac{4}{3} + \frac{1}{m}

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D

\frac{S_{A}}{S_{B}} = \frac{1}{3} + \frac{1}{m}

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Solution

The correct option is A $\frac{S_{A}}{S_{B}} = 2 + \frac{1}{m}$
$S_{A} = 2 m + (2 m + 1) + (2 m + 2) + . . . + 4 m$
Number of terms in the series $S_{A}$ is $2 m + 1$
$S_{A} = \frac{2 m + 1}{2} (4 m + 2 m) ∴ S_{A} = 3 m (2 m + 1)$

$S_{B} = (2 m + 1) + (2 m + 3) + . . . + (4 m - 1)$
Number of terms in the series $S_{B}$ is $m$
$S_{B} = \frac{m}{2} [2 m + 1 + 4 m - 1] ∴ S_{B} = 3 m^{2}$

Therefore,
$\frac{S_{A}}{S_{B}} = \frac{2 m + 1}{m} = 2 + \frac{1}{m}$

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Arithmetic Mean

Standard XII Mathematics

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