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mathmath333

  • one year ago

Fun question

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{Let }\ a,b,c \ \normalsize \text{be positive integers such that} \hspace{.33em}\\~\\ & \dfrac{b}{a}\ \normalsize \text{is an integer }\hspace{.33em}\\~\\ & \normalsize \text{if } \ a,b,c \ \ \normalsize \text{are in geometric progression. } \hspace{.33em}\\~\\ & \normalsize \text{and the arithmetic mean of } \ a,b,c \ \ \normalsize \text{is}\ \ (b+2) \hspace{.33em}\\~\\ & \normalsize \text{then find the value of }\ \left(\dfrac{a^2+a-14}{a+1} \right)\hspace{.33em}\\~\\ \end{align}}\)

  2. Afrodiddle
    • one year ago
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    It doesn't look fun

  3. anonymous
    • one year ago
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    agreed @Afrodiddle

  4. imqwerty
    • one year ago
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    4

  5. mathmath333
    • one year ago
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    correct, how u got that

  6. imqwerty
    • one year ago
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    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

  7. ganeshie8
    • one year ago
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    Nice!

  8. Isaiah.Feynman
    • one year ago
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    I solved it halfway in my head.

  9. mathmath333
    • one year ago
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    nice

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