Sequence-And-Series Question 673
Question: a, b, c are the first three terms of a geometric series. If the harmonic mean of a and b is 12 and that of b and c is 36, then the first five terms of the series are
Options:
A) 9, 18, 27, 36, 45
B) 8, 24, 72, 216, 648
C) 4, 22, 38, 46
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Let $ b=ar,c=ar^{2} $ Given that $ 12=\frac{2ab}{a+b}=\frac{2a.ar}{a+ar}=\frac{2ar}{1+r} $ or $ ar=6(1+r) $ ?..(1) Also, $ 36=\frac{2bc}{b+c}=\frac{2.ar.ar^{2}}{ar+ar^{2}}=\frac{2ar^{2}}{1+r} $
$ \Rightarrow ar^{2}=18(1+r) $ ?..(2) Dividing (ii) by (i), we have $ \frac{ar^{2}}{ar}=\frac{18(1+r)}{6(1+r)}\Rightarrow r=3 $
$ \therefore $ From (i), $ a\times 3=6(1+3)\Rightarrow a=\frac{6\times 4}{3}=8 $
$ \therefore $ First five numbers are $ 8,24,72,216,648. $
BETA
