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2013-05-29T23:42:27-03:00

\frac{3a}{2}-a=\ \frac{3a-2a}{2}=\frac a{2}\ \ logo\ a\ razao\ r=\frac{a}{2}

 

agora calculamos o 10º termo assim:

 

a_n=a_1+(n-1)r\\a_1_0=a+(10-1)*\frac{a}{2}\\a_1_0=a+\frac{9a}{2}\\a_1_0=\frac{11a}{2}

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2013-05-30T17:53:34-03:00

Temos que:

 

\text{a}+\text{r}=\dfrac{3\text{a}}{2}

 

Logo:

 

\text{r}=\dfrac{3\text{a}}{2}-\text{a}=\dfrac{3\text{a}-2\text{a}}{2}=\dfrac{\text{a}}{2}

 

Observe que:

 

\text{a}_{\text{n}}=\text{a}_1+(\text{n}-1)\cdot\text{r}

 

\text{a}_{10}=\text{a}+(10-1)\cdot\frac{\text{a}}{2}

 

\text{a}_{10}=\text{a}+9\cdot\dfrac{\text{a}}{2}

 

\text{a}_{10}=\text{a}+\dfrac{9\text{a}}{2}

 

\text{a}_{10}=\dfrac{2\text{a}+9\text{a}}{2}=\dfrac{11\text{a}}{2}

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