Sequence And Series Question 406
Question: There are four numbers of which the first three are in G.P. and the last three are in A.R, whose common difference is 6. If the first and the last numbers are equal then two other numbers are
Options:
A) -2, 4
B) -4, 2
C) 2, 6
D) none
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Let the last three numbers in A.P. be a, $ a+6, $ $ a+12, $ then the first term is also $ a+12, $ . But $ a+12, $ a, $ a+6 $ are in GP.
$ \therefore a^{2}=(a+12),(a+6)\Rightarrow a^{2}=a^{2}+18a+72 $
$ \therefore a=-4. $
$ \therefore $ The numbers are $ 8,-4,2,8. $
BETA
