If the first,second and last terms of an A.P. | vedclass

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(C) Given,the first term $A = a$. The second term is $A + d = b$,where $d$ is the common difference. Thus,$d = b - a$. The last term $l = 2a$. The for

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$\frac{3ab}{2(b - a)}$

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(C) Given,the first term $A = a$. The second term is $A + d = b$,where $d$ is the common difference. Thus,$d = b - a$. The last term $l = 2a$. The formula for the $n^{th}$ term is $l = A + (n - 1)d$. Substituting the values: $2a = a + (n - 1)(b - a)$. $a = (n - 1)(b - a) \implies n - 1 = \frac{a}{b - a}$. $n = \frac{a}{b - a} + 1 = \frac{a + b - a}{b - a} = \frac{b}{b - a}$. The sum of an $A.P.$ is given by $S_n = \frac{n}{2}(A + l)$. Substituting $n = \frac{b}{b - a}$,$A = a$,and $l = 2a$: $S_n = \frac{b}{2(b - a)}(a + 2a) = \frac{b}{2(b - a)}(3a) = \frac{3ab}{2(b - a)}$.

If the first,second and last terms of an $A.P.$ are $a, b, 2a$ respectively,then its sum will be

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