# 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

But try the problem first...

Answer: \(1\)

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

## Other useful links

- https://www.cheenta.com/probability-in-divisibility-amc-10a-2003-problem-15/
- https://www.youtube.com/watch?v=Ch92mPHnl-c