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

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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