First slide

Means - A.M/G.M/H.M

Question

$n A.M's are inserted between n and 3 n + 2 then 10^{th} A.M. is$

Moderate

A

$n + 19$
B

$n + 20$
C

$n + 21$
D

$n + 22$

Solution

$A_{r} = a + r (\frac{b - a}{n + 1}) \Rightarrow A_{10} = n + 10 (\frac{3 n - 2 - n}{n + 1}) = n + 20$

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