Question
Question: Let b<sub>1</sub>,b<sub>2</sub>,b<sub>3</sub> (b<sub>1</sub> > 0) are three consecutive terms of a G...
Let b1,b2,b3 (b1 > 0) are three consecutive terms of a GP, with common ratio r. If b3 > 4b2 – 3b1, then r may be :
A
r < 1
B
r > 3
C
(1) or (2)
D
1 < r < 3
Answer
(1) or (2)
Explanation
Solution
Q b2 = rb1 & b3 = r2b1
Q b3 > 4b2 – 3b1
Ž r2 > 4r – 3
Ž r2 – 4r + 3 > 0
Ž (r – 1) (r – 3)
\ r < 1 or r > 3