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# Geometric Progression- ISI Entrance B. Stat (Hons) 2003- Problem 3

## Geometric Progression

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.

## Try the problem

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) $-1<x<3 ;$

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

Geometric Progression

6 out of 10

challenges and thrills of pre college mathematics

## Use some hints

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 $b= \sqrt{ac}$.

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.

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