INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More

Content

[hide]

A geometric progression is a sequence in which each term is derived by multiplying or dividing the preceding term by a fixed number called the common ratio. E.g., the height to which a ball rises in each successive bounce follows a geometric progression. The sequence 4, -2, 1,... is a Geometric Progression (GP) for which (-1/2) is the common ratio. We can use the concept to find an arbitrary term, a finite or infinite sum of the series, and apply them in various contexts, including some difficult problems.

Suppose that the three distinct real numbers \(a,b \text{ and } c\) are in G.P. and \(a+b+c=xb\). Then

(A) \(-3<x<1 ;\)

(B) \(x>1\) or \(x<-3 ;\)

(C) \(x>3\) or \(x<-1 ;\)

(D) \(-1<x<3 ;\)

Source

Competency

Difficulty

Suggested Book

ISI entrance B. Stat. (Hons.) 2003 problem 3

Geometric Progression

6 out of 10

challenges and thrills of pre college mathematics

First hint

If any three quantity are in GP then we have a relation between them, in this case \(a,b,c\) are in G.P. so we have

\(b^2=ac\) or \(b= \sqrt{ac}\).

Second Hint

We also have \(a+b+c=xb\) so

which is equal to \(a+c=b(x-1)\)

=\(\frac{a}{b}+\frac{c}{b}=(x-1)\)

since \(b=\sqrt{ac}\) we will get

=\(\sqrt{\frac{a}{c}}+\sqrt{\frac{c}{a}}=(x-1)\)

=\(\sqrt{\frac{a}{c}}+\frac{1}{\sqrt{\frac{a}{c}}}=(x-1)\)

Let \(\sqrt{\frac{a}{c}}=k\), then we have the form of \(k+\frac{1}{k}\) which we know has a value either greater than 2 or less than -2.

so we can write

either, \(x-1 > 2\) or \(x-1 < -2\)

and now you can easily get the answer.

Final Step

So the answer is x>3 or x<-1.

- https://www.cheenta.com/problem-based-on-divisibility-cmi-2015-problem-3/
- https://www.youtube.com/watch?v=7Zx5n3nuGmo

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

JOIN TRIAL
Google