Try this beautiful problem from Algebra : AP & GP from AMC 10A, 2004.
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?
Algebra
AP
GP
Answer: \(1\)
AMC-10A (2003) Problem 18
Pre College Mathematics
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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