mathmath333
  • mathmath333
Fun question
Mathematics
chestercat
  • chestercat
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{Let }\ a,b,c \ \normalsize \text{be positive integers such that} \hspace{.33em}\\~\\ & \dfrac{b}{a}\ \normalsize \text{is an integer }\hspace{.33em}\\~\\ & \normalsize \text{if } \ a,b,c \ \ \normalsize \text{are in geometric progression. } \hspace{.33em}\\~\\ & \normalsize \text{and the arithmetic mean of } \ a,b,c \ \ \normalsize \text{is}\ \ (b+2) \hspace{.33em}\\~\\ & \normalsize \text{then find the value of }\ \left(\dfrac{a^2+a-14}{a+1} \right)\hspace{.33em}\\~\\ \end{align}}\)
Afrodiddle
  • Afrodiddle
It doesn't look fun
anonymous
  • anonymous

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imqwerty
  • imqwerty
4
mathmath333
  • mathmath333
correct, how u got that
imqwerty
  • imqwerty
b/a = r(integer) a=a b=ar c=ar^2 (a+ar+ar^2)/3 = ar+2 solving we get - r^2 + r(-2)+(1-6/r) = 0 solve for r we get (- 2 +- (4-4+24/a)^(1/2))/2 1+- 1/2(24/a)^(1/2) well r = integer therefore 24/a = perfect square therefore a=6 we need to find (a^2 + a-14)/(a+1) plug a = 6 and get answer = 4
ganeshie8
  • ganeshie8
Nice!
Isaiah.Feynman
  • Isaiah.Feynman
I solved it halfway in my head.
mathmath333
  • mathmath333
nice

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