Sequence And Series Question 270
Question: If the product of three terms of G.P. is 512. If 8 added to first and 6 added to second term, so that number may be in A.P., then the numbers are
[Roorkee 1964]
Options:
A) 2, 4, 8
B) 4, 8, 16
C) 3, 6, 12
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let three terms of a G.P. are $ \frac{a}{r},\ a,\ ar $ So $ \frac{a}{r}.\ a.\ ar=512 $
$ \Rightarrow $ $ a^{3}=8^{3} $
$ \Rightarrow $ $ a=8 $ From second condition, we get $ \frac{a}{r}+8,\ a+6,\ ar $ will be in A.P.
$ \Rightarrow $ $ 2(a+6)=\frac{a}{r}+8+ar $
$ \Rightarrow $ $ 28=8{ \frac{1}{r}+1+r } $
$ \Rightarrow $ $ \frac{1}{r}+r+1=\frac{7}{2} $
$ \Rightarrow $ $ \frac{1}{r}+r-\frac{5}{2}=0 $
$ \Rightarrow $ $ r^{2}-\frac{5}{2}r+1=0 $
$ \Rightarrow $ $ 2r^{2}-5r+2=0 $
$ \Rightarrow $ $ (2r-1)(r-2)=0 $
$ \Rightarrow $ $ r=\frac{1}{2},\ r=2 $ $ (\because \ r>1) $
$ \Rightarrow $ $ r=2 $ . Hence required numbers are $ 4,\ 8,\ 16 $ . Trick: Check for (a) $ 2+8,\ 4+6,\ 8 $ are not in A.P. (b) $ 4+8,\ 8+6,\ 16\ i.e.\ 12,\ 14,\ 16 $ are in A.P.
BETA
