If the number $(2 n - 1), (3 n + 2)$ and $(6 n - 1)$ are in AP, find the value of $n$ and the numbers.

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Solution

It is given that the numbers $(2 n - 1), (3 n + 2)$ and $(6 n - 1)$ are in AP.

Therefore, $(3 n + 2) - (2 n - 1) = (6 n - 1) - (3 n + 2)$

$=> 3 n + 2 - 2 n + 1 = 6 n - 1 - 3 n - 2$

$=> n + 3 = 3 n - 3$

$=> 2 n = 6$

$=> n = 3$

When $n = 3,$

$2 n - 1 = 2$ $\times$ $3 - 1 = 6 - 1 = 5$

$3 n + 2 = 3$ $\times$ $3 + 2 = 9 + 2 = 11$

$6 n - 1 = 6$ $\times$ $3 - 1 = 18 - 1 = 17$

Hence, the required value of n is 3 and the numbers are 5, 11 and 17.

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