જો x1, x2, ……, xn નો મધ્યક overlinex હોય...

જો `x_1, x_2, ……, x_n` નો મધ્યક `overlinex` હોય, તો `a ne 0`, માટે `ax_1, ax_2, …….., ax_n`, `x_1/a, x_2/a, …….., x_n/a` નો મધ્યક = ........

A

`(a+1/a)overlinex`

B

`(a+1/a)overlinex/2`

C

`(a+1/a)overlinex/n`

D

`((a+1/a)overlinex)/(2n)`

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