# AP GP Problem | AMC-10A, 2004 | Question 18

Try this beautiful problem from Algebra : AP & GP from AMC 10A, 2004.

## AP GP - AMC-10A, 2004- Problem 18

A sequence of three real numbers forms an arithmetic progression with a first term of $$9$$. If $$2$$ is added to the second term and $$20$$ is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term in the geometric progression?

• $$1$$
• $$4$$
• $$36$$
• $$49$$
• $$81$$

Algebra

AP

GP

## Check the Answer

Answer: $$1$$

AMC-10A (2003) Problem 18

Pre College Mathematics

## Try with Hints

We assume the common difference in the AP series is $$d$$ .....Therefore the numbers will be $$9, (9+d),(9+2d)$$ .Therefore according to the condition if we add $$2$$ with $$2$$nd term and add $$20$$ to the third term the numbers becomes in Geometric Progression........$$9$$ , $$(9+d+2)=11+d$$ , $$(9+2d+20)=29+2d$$

can you finish the problem........

Now according to the Geometric Progression , $$\frac{11+d}{9}=\frac{29+d}{11+d}$$

$$\Rightarrow (11+d)^2 =9(29+2d)$$

$$\Rightarrow d^2 +4d-140=0$$

$$\Rightarrow (d+14)(d-10)=0$$

$$\Rightarrow 10 ,-14$$

can you finish the problem........

Therefore we choose the value of $$d=-14$$ (as smallest possible value for the third term)

The third term will be $$2(-14)+29=1$$

# GP and 2-digit number | PRMO 2017 | Question 16

Try this beautiful problem from the Pre-RMO, 2017 based on GP and 2-digit number.

## GP and 2-digit number - PRMO 2017

Five distinct 2-digit numbers are in geometric progression. Find the middle term.

• is 107
• is 36
• is 840
• cannot be determined from the given information

### Key Concepts

Geometric Progression

2-digit number

Middle term

## Check the Answer

PRMO, 2017, Question 16

Elementary Algebra by Hall and Knight

## Try with Hints

First hint

Let the numbers be $$a, ar, ar^{2}, ar^{3}, ar^{4}$$

for all are two digit number r=$$\frac{2}{3}$$ or $$\frac{3}{2}$$

Second Hint

where fourth power of integers greater than 3 are 3 digit numbers

Final Step

then five numbers (16,24,36,54,81)

then middle term=36.

# Problem on Geometric Progression | PRMO 2017 | Question 14

Try this beautiful problem from the Pre-RMO, 2017 based on Geometric Progression.

## Problem on Geometric Progression - PRMO 2017

Suppose x is positive real number such that {x},[x] and x are in geometric progression. Find the least positive integer n such that $$x^{n} \gt 100$$ where [x] denotes the integer part of x and {x} =x -[x]

• is 107
• is 10
• is 840
• cannot be determined from the given information

### Key Concepts

Geometric Progression

Greatest Integer

Real Number

## Check the Answer

PRMO, 2017, Question 14

Elementary Algebra by Hall and Knight

## Try with Hints

First hint

here we have $$[x]^{2}$$=x{x}

$$\Rightarrow$$ {x}=a, [x]=ar, $$x=ar^{2}$$

$$\Rightarrow a+ar=ar^{2}$$

$$\Rightarrow r^{2}-r-1=0$$

$$\Rightarrow r=\frac{1+\sqrt{5}}{2}$$

Second Hint

Let ar=I

$$\Rightarrow a=\frac{2I}{1+\sqrt{5}}=\frac{I(\sqrt{5}-1)}{2}$$

for 0 $$\lt$$ a $$\lt$$ 1 $$\Rightarrow 0 \lt \frac{I(\sqrt{5}-1)}{2} \lt 1$$

$$\Rightarrow 0 \lt I \lt \frac{(\sqrt{5}+1)}{2}$$

Final Step

$$\Rightarrow$$ I=1

$$\Rightarrow$$ ar=1

$$\Rightarrow a=\frac{2}{\sqrt{5}+1}=\frac{\sqrt{5}-1}{2}$$

$$x=ar^{2}=r=\frac{\sqrt{5}+1}{2}$$

$$\Rightarrow (\frac{\sqrt{5}+1}{2})^{n} \gt 100$$

$$\Rightarrow Nlog_{10}(\frac{\sqrt{5}+1}{2}) \gt 2$$

$$\Rightarrow N \gt 9.5$$

$$\Rightarrow N_{min}$$=10.

# RMO 2018 Tamil Nadu Problem 2 - real quadratic

RMO 2018 Tamil Nadu Problem 2 is from Theory of Equations. We present sequential hints for this problem. Do not read all hints at one go. Try it yourself.

# Problem

Find the set of all real values of a for which the real polynomial equation $P(x) = x^2 - 2ax + b = 0$ has real roots, given that $P(0) \cdot P(1) \cdot P(2) \neq 0$ and $( P(0), P(1), P(2) )$ form a geometric progression.

## Key ideas you will need to solve this problem

• Theory of Equations:
• Discriminant of a quadratic is non -negative iff roots are real.
• If a, b, c are in Geometric Progression then $ac = b^2$

Also see

## Hint 1: Compute P(0), P(1), P(2)

P(0) = b (plug in x = 0 in the given quadratic). P(1) = 1 - 2a + b and P(2) = 4 - 4a + b.

Since P(0), P(1) and P(2) are in Geometric Progression, hence $P(0) \times P(2) = P(1)^2$. This balls down to $$b \cdot (4 - 4a + b) = (1 - 2a + b)^2$$. Simplify this expression.

## Hint 2: Express b in terms of a

$b \cdot (4 - 4a + b) = (1 - 2a + b)^2 \ \Rightarrow 4b - 4ab + b^2 = 1 + 4a^2 + b^2 - 4a - 4ab + 2b \ \Rightarrow 2b = 4a^2 - 4a + 1$

## Hint 3: Use the discriminant of the quadratic

Discriminant of a quadratic equation $ax^2 + bx + c = 0$ is $b^2 - 4ac$. The quadratic equation has real roots iff $b^2 - 4ac \geq 0$.

The discriminant of the given quadratic is $$4a^2 - 4b$$ From Hint 2, we know $$4b = 8a^2 - 8a + 2$$ Hence the discriminant is $$4a^2 - (8a^2 - 8a + 2) = -4a^2 + 8a - 2$$

We want this discriminant to be non negative.

## Hint 4: Final Lap

$$-4a^2 + 8a - 2 \geq 0 \Rightarrow 2a^2 - 4a + 1 \leq 0$$

Use the quadratic formula to find the roots of the quadratic $2a^2 - 4a + 1$(this will allow us to factorize the expression).

The roots are $$\frac { 4 \pm \sqrt {16 - 8}}{2} = 2 \pm \sqrt {2}$$

Hence $a \in [2 - \sqrt 2, 2 + \sqrt 2]$

# Reference:

• These ideas are usually discussed in the Polynomials I module of Cheenta Math Olympiad Program.
• Challenges and Thrills of Pre -College Mathematics by Venkatchala is a good reference for some these ideas.

Also see

RMO 2018 Tamil Nadu Region