Sequence And Series Question 629
Question: If $ x $ is added to each of numbers 3, 9, 21 so that the resulting numbers may be in G.P., then the value of $ x $ will be
[MP PET 1986]
Options:
A) 3
B) $ \frac{1}{2} $
C) 2
D) $ \frac{1}{3} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ 3+x,\ 9+x,\ 21+x $ are in G.P.
$ \Rightarrow $ $ {{(9+x)}^{2}}=(3+x)(21+x) $
$ \Rightarrow $ $ 81+x^{2}+18x=x^{2}+24x+63 $
$ \Rightarrow $ $ 6x=18 $ or $ x=3 $ . Trick: Check for (a), $ 3+3,\ 9+3,\ 21+3 $ are in G.P.
BETA
