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(A) Let the first term be $a$,the second term be $b$,and the last term be $c$.The common difference $d = b - a$.The formula for the $n^{th}$ term of a

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$\frac{b + c - 2a}{b - a}$

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(A) Let the first term be $a$,the second term be $b$,and the last term be $c$.The common difference $d = b - a$.The formula for the $n^{th}$ term of an arithmetic progression is $l = a + (n - 1)d$,where $l$ is the last term.Substituting the values,we get $c = a + (n - 1)(b - a)$.Rearranging the terms to solve for $n$:$c - a = (n - 1)(b - a)$$n - 1 = \frac{c - a}{b - a}$$n = \frac{c - a}{b - a} + 1$$n = \frac{c - a + b - a}{b - a}$$n = \frac{b + c - 2a}{b - a}$.

If $a$,$b$,and $c$ are the first,second,and last terms of an arithmetic progression respectively,then the total number of terms is......

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