Question

1. Interpolar 5 meios aritméticos entre -2 e 40.
2. Quando inserimos 10 meios aritméticos entre 2 e 79, qual é a razão da PA obtida?
3. Inscrevendo-se nove meios aritméticos entre 15 e 45, qual é o sexto termo da PA?

me ajudaaa

Answer 1

Explicação passo-a-passo:

1)

$\sf a_n=a_1+(n-1)\cdot r$

$\sf a_7=a_1+6r$

$\sf 40=-2+6r$

$\sf 6r=40+2$

$\sf 6r=42$

$\sf r=\dfrac{42}{6}$

$\sf \red{r=7}$

$\sf \red{PA(-2,5,12,19,26,33,40)}$

2)

$\sf a_n=a_1+(n-1)\cdot r$

$\sf a_{12}=a_1+11r$

$\sf 79=2+11r$

$\sf 11r=79-2$

$\sf 11r=77$

$\sf r=\dfrac{77}{11}$

$\sf \red{r=7}$

3)

$\sf a_n=a_1+(n-1)\cdot r$

$\sf a_{11}=a_1+10r$

$\sf 45=15+10r$

$\sf 10r=45-15$

$\sf 10r=30$

$\sf r=\dfrac{30}{10}$

$\sf r=3$

O sexto termo é:

$\sf a_6=a_1+5r$

$\sf a_6=15+5\cdot3$

$\sf a_6=15+15$

$\sf \red{a_6=30}$

Answer 2

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$\Large\boxed{\sf{\underline{Interpolac_{\!\!,}\tilde{a}o~Aritm\acute{e}tica}}}\\\sf{Consiste~em~inserir~termos~entre~dois~n\acute{u}meros~dados}\\\sf{de~modo~a~obter~uma~PA}$

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1)

$\sf{a_1=-2~~a_7=40~~r=?}\\\sf{a_7=a_1+6r}\\\sf{40=-2+6r}\\\sf{6r=40+2}\\\sf{6r=42}\\\sf{r=\dfrac{42}{6}}\\\sf{r=7}\\\tt{A~PA~ser\acute{a}:}\\\huge\boxed{\boxed{\boxed{\sf{(\underline{-2},5,12,19,26,33,\underline{40})}}}}$

2)

$\sf{a_1=2~~a_{12}=79~~r=?}\\\sf{a_{12}=a_1+11r}\\\sf{79=2+11r}\\\sf{11r=79-2}\\\sf{11r=77}\\\sf{r=\dfrac{77}{11}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf{r=7}}}}}$

3)

$\sf{a_1=15~~a_{11}=45~~r=?~~a_6=?}\\\sf{a_{11}=a_1+10r}\\\sf{45=15+10r}\\\sf{10r=45-15}\\\sf{1\diagup\!\!\!0r=3\diagup\!\!\!0\implies r=3}\\\sf{a_6=a_1+5r}\\\sf{a_6=15+5\cdot3}\\\sf{a_6=15+15}\\\huge\boxed{\boxed{\boxed{\boxed{\sf{a_6=30\checkmark}}}}}$