Sequence And Series Question 318
Question: If the first, second and last terms of an A.P. be $ a,\ b,\ 2a $ respectively, then its sum will be
Options:
A) $ \frac{ab}{b-a} $
B) $ \frac{ab}{2(b-a)} $
C) $ \frac{3ab}{2(b-a)} $
D) $ \frac{3ab}{4(b-a)} $
Show Answer
Answer:
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).
BETA
