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 :
We have first term \[A=a\] ......(i) Second term \[A+d=b\] ......(ii) and last term \[l=2a\] ......(iii) From (i), (ii) and (iii), \[\Rightarrow \] and \[n=\frac{b}{b-a}\] Then sum \[S=\frac{n}{2}[a+l]=\frac{b}{2(b-a)}[a+2a]=\frac{3ab}{2(b-a)}\] Trick: Let \[a=2,\ b=3\]then the sum \[=9\] which is given by option (c).
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