The arithmetic mean of the set of variable a, a+d, a + 2d, a + 3d...a + 2nd is

a + nd

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a - nd

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(a + n)d

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ad + n

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Solution

The correct option is A a + nd
In series a,a+d,a+2d..........a+2nd
The first term =a , common difference=d
Term =2n+1
Last term =a+2nd
We know that sum of series

$= \frac{2 n + 1}{2} (f i s t t e r m + l a s t t e r m)$

$= \frac{2 n + 1}{2} (a + a + 2 n d) = (2 n + 1) (a + n d)$

Then mean of series

$= \frac{sum of series}{2 n + 1}$

$= \frac{(2 n + 1) (a + n d)}{n} = (a + n d)$

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