A) \[\frac{ab}{b-a}\]
B) \[\frac{ab}{2(b-a)}\]
C) \[\frac{3ab}{2(b-a)}\]
D) \[\frac{3ab}{4(b-a)}\]
Correct Answer: C
Solution :
[c] : We have first term = a ...(i) Second term = b ...(ii) and last term = 2a ...(iii) Let d be the common difference. From (i), (ii) and (iii), d = (b - a) and \[n=\frac{b}{b-a}\]. Then, sum \[(S)=\frac{n}{2}[a+l]\] \[=\frac{b}{2(b-a)}[a+2a]=\frac{3ab}{2(b-a)}\]You need to login to perform this action.
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