In the formulae S(n)=(n)/(2){2a+(n-1)d} ...

In the formulae `S_(n)=(n)/(2){2a+(n-1)d}` make d as the subject
The following steps are involved in solving the above problem. Arrange them in sequential order.
(A) `(n-1)d=(2S_(n))/(n)-2a`
(B) Given, `S_(n)=(n)/(2)[2a+(n-1)d]rArrn[2a+(n-1)d]`
=`2S_(n)`
(C) `rArr d=(2)/(n-1)[(S_(n))/(n)-a]`
(D) `2a+(n-1)d=(2S_(n))/(n)`

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