Sequence And Series Question 293
Question: If first three terms of sequence $ \frac{1}{16},a,b,\frac{1}{6} $ are in geometric series and last three terms are in harmonic series, then the value of $ a $ and $ b $ will be
[UPSEAT 1999]
Options:
A) $ a=-\frac{1}{4},b=1 $
B) $ a=\frac{1}{12},b=\frac{1}{9} $
C) (a) and (b) both are true
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
If $ \frac{1}{16},a,b $ are in G.P., then $ a^{2}=\frac{b}{16} $ or $ 16,a^{2}=b $ …..(i) and if $ a,b,\frac{1}{6} $ are in H.P., then $ b=\frac{2.a.\frac{1}{6}}{a+\frac{1}{6}}=\frac{2a}{6a+1} $ ?..(ii) From (i) and (ii), $ 16a^{2}=\frac{2a}{6a+1} $ or $ 2a( 8a-\frac{1}{6a+1} )=0 $ or $ 8a,(6a+1)-1=0 $ or $ 48a^{2}+8a-1=0 $ , $ (\because a\ne 0) $ or $ (4a+1)(12a-1)=0 $
$ \therefore $ $ a=-\frac{1}{4},\frac{1}{12} $ When $ a=-\frac{1}{4} $ then from (i), $ b=16,{{( -\frac{1}{4} )}^{2}}=1 $ When $ a=\frac{1}{12} $ then from (i), $ b=16{{( \frac{1}{12} )}^{2}}=\frac{1}{9} $ Therefore, $ a=-\frac{1}{4},b=1 $ or $ a=\frac{1}{12},b=\frac{1}{9} $ .
BETA
