Os números 3, a e b estão em progressão aritmética crescente.Os números 3, (a+1) e (b+5) estão em progressão geométrica crescente. O valor de a+b é:
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P.A.(3,a,b)
Assim temos:
![a=\frac{3+b}{2}\\\\
2a=3+b\\\\
\boxed{b=2a-3} a=\frac{3+b}{2}\\\\
2a=3+b\\\\
\boxed{b=2a-3}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B3%2Bb%7D%7B2%7D%5C%5C%5C%5C%0A2a%3D3%2Bb%5C%5C%5C%5C%0A%5Cboxed%7Bb%3D2a-3%7D)
P.G.(3,a+1,b+5)
Assim temos:
![\frac{a+1}{3}=\frac{b+5}{a+1}\\\\
Vimos \ anteriormente \ que \ b=2a-3, \ assim:\\\\
\frac{a+1}{3}=\frac{(2a-3)+5}{a+1}\\\\
\frac{a+1}{3}=\frac{2a+2}{a+1}\\\\
\frac{a+1}{3}=\frac{2(a+1)}{a+1}\\\\
\frac{a+1}{3}=2\\\\
a+1=2.3\\
a+1=6\\
a=5\\\\
Como \ b=2a-3, \ temos:\\
b=2.5-3=10-3\\
b=7\\\\
Calculando \ a+b:\\
a+b=5+7=12
\frac{a+1}{3}=\frac{b+5}{a+1}\\\\
Vimos \ anteriormente \ que \ b=2a-3, \ assim:\\\\
\frac{a+1}{3}=\frac{(2a-3)+5}{a+1}\\\\
\frac{a+1}{3}=\frac{2a+2}{a+1}\\\\
\frac{a+1}{3}=\frac{2(a+1)}{a+1}\\\\
\frac{a+1}{3}=2\\\\
a+1=2.3\\
a+1=6\\
a=5\\\\
Como \ b=2a-3, \ temos:\\
b=2.5-3=10-3\\
b=7\\\\
Calculando \ a+b:\\
a+b=5+7=12](https://tex.z-dn.net/?f=%5Cfrac%7Ba%2B1%7D%7B3%7D%3D%5Cfrac%7Bb%2B5%7D%7Ba%2B1%7D%5C%5C%5C%5C%0AVimos+%5C+anteriormente+%5C+que+%5C+b%3D2a-3%2C+%5C+assim%3A%5C%5C%5C%5C%0A%5Cfrac%7Ba%2B1%7D%7B3%7D%3D%5Cfrac%7B%282a-3%29%2B5%7D%7Ba%2B1%7D%5C%5C%5C%5C%0A%5Cfrac%7Ba%2B1%7D%7B3%7D%3D%5Cfrac%7B2a%2B2%7D%7Ba%2B1%7D%5C%5C%5C%5C%0A%5Cfrac%7Ba%2B1%7D%7B3%7D%3D%5Cfrac%7B2%28a%2B1%29%7D%7Ba%2B1%7D%5C%5C%5C%5C%0A%5Cfrac%7Ba%2B1%7D%7B3%7D%3D2%5C%5C%5C%5C%0Aa%2B1%3D2.3%5C%5C%0Aa%2B1%3D6%5C%5C%0Aa%3D5%5C%5C%5C%5C%0AComo+%5C++b%3D2a-3%2C+%5C+temos%3A%5C%5C%0Ab%3D2.5-3%3D10-3%5C%5C%0Ab%3D7%5C%5C%5C%5C%0ACalculando+%5C+a%2Bb%3A%5C%5C%0Aa%2Bb%3D5%2B7%3D12%0A)
Assim temos:
P.G.(3,a+1,b+5)
Assim temos:
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