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?

  • \(1\)
  • \(4\)
  • \(36\)
  • \(49\)
  • \(81\)

Key Concepts

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 \( 2(-14)+29=1\)

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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

Answer: is 36.

PRMO, 2017, Question 16

Elementary Algebra by Hall and Knight

Try with Hints

Let the numbers be \(a, ar, ar^{2}, ar^{3}, ar^{4}\)

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

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

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

then middle term=36.

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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

Answer: is 10.

PRMO, 2017, Question 14

Elementary Algebra by Hall and Knight

Try with Hints

here we have \([x]^{2}\)=x{x}

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

\(\Rightarrow a+ar=ar^{2}\)

\(\Rightarrow r^{2}-r-1=0\)

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

Let ar=I

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

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

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

\(\Rightarrow\) 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.

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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

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Advanced Math Olympiad Program

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:

Also see

RMO 2018 Tamil Nadu Region