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# AP GP Problem | AMC-10A, 2004 | Question 18

Try this beautiful problem from Algebra based on AP GP from AMC-10A, 2004. You may use sequential hints to solve the problem.

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?

• $1$
• $4$
• $36$
• $49$
• $81$

Algebra

AP

GP

## Check the Answer

Answer: $1$

AMC-10A (2003) Problem 18

Pre College Mathematics

## Try with Hints

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 $2(-14)+29=1$