Question

Mais uma de PA

O 10º termo da PA (a,3a/2,...) é igual a:

Answer 1

$<var>\frac{3a}{2}-a=\ \frac{3a-2a}{2}=\frac a{2}\ \ logo\ a\ razao\ r=\frac{a}{2}</var>$

 

agora calculamos o 10º termo assim:

 

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

Answer 2

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:

 

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

 

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

 

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

 

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

 

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