The mean of the series $a, a + d, a + 2 d, . . ., a + 2 n d$ is

a + (n - 1) d

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a + n d

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a + (n + 1) d

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None of these

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Solution

The correct option is A $a + n d$
Required mean $= \frac{a + (a + d) + (a + 2 d) + . . . + (a + 2 n d)}{2 n + 1}$
$= \frac{\frac{2 n + 1}{2} {2 a + (2 n + 1 - 1) d}}{2 n + 1} = a + n d$

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Similar questions

a, a + d, a + 2 d + . . . . . ., a + (2 n - 1) d, a + 2 n d

a

n

d

(2 n - 1) d = 10.5 . . . . (i)

n (a + (n - 1) d) = 2 d . . . . . . (i i)

n (a + n d) = 30 . . . . . . . (i i i)

\frac{\sqrt{a} + \sqrt{a + n d}}{\sqrt{a} + \sqrt{a + d}} + \frac{\sqrt{a} + \sqrt{a + n d}}{\sqrt{a + d} + \sqrt{a + 2 d}} + \dots + \frac{\sqrt{a} + \sqrt{a + n d}}{\sqrt{a + (n - 1) d} + \sqrt{a + n d}}

\frac{n}{2} [2 a + (n - 1) d]

$a + (a + d) + (a + 2 d) + . . . . + (a + (n - 1) d) = \frac{n}{2} [2 a + (n - 1) d]$

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