Q 2908680508.     `S =\frac{n}{2} [ 2a+ ( n-1 )d ]`; make 'd' the subject of formula

A

`d=\frac{ ( 2S-a )}{n ( n-1 )}`

B

`d=\frac{ ( 2S-na )}{n ( n+1 )}`

C

`d=\frac{2 ( S-na )}{n ( n-1 )}`

D

`d=\frac{2 ( S-a )}{n ( n+1 )}`

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