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


But try the problem first…

Answer: \(1\)

Source
Suggested Reading

AMC-10A (2003) Problem 18

Pre College Mathematics

Try with Hints


First hint

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……..

Second Hint

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……..

Final Step

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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