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

Try this beautiful problem from Algebra based on AP GP from AMC-10A, 2004. You may use sequential hints to solve the problem.

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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