If the first, second and last term of an A.P. are $x, y$ and $2 x$ respectively, its sum is

\frac{x y}{2 (y - x)}

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\frac{3 x y}{2 (y - x)}

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\frac{2 x y}{y - x}

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\frac{x y}{y - x}

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Solution

The correct option is B $\frac{3 x y}{2 (y - x)}$
Here $a_{1} = x,$ $a_{2} = y$ and $a_{n} = 2 x$ .
$∴$ First term, $a = x$ and common difference, $d = y - x$
Since, $a_{n} = a + (n - 1) d$
$\Rightarrow 2 x = x + (n - 1) (y - x)$
$\Rightarrow x = (n - 1) (y - x)$
$\Rightarrow \frac{x}{y - x} = n - 1$
$\Rightarrow \frac{x}{y - x} + 1 = n$
$\Rightarrow n = \frac{y}{y - x}$

Now, we know that, $S_{n} = \frac{n}{2} (a + a_{n})$

$∴ S_{n} = \frac{\frac{y}{y - x}}{2} (x + 2 x)$

$\Rightarrow S_{n} = \frac{\frac{y}{y - x}}{2} (3 x)$

$\Rightarrow S_{n} = \frac{3 x y}{2 (y - x)}$

Hence, option B is correct.

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