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