Question

interpole doze meios ariméticos entre $\frac{-3}{4}$ e$\frac{11}{6}$

Answer 1

$(a_1,\underbrace{a_2,a_3,a_4,\dots,a_{12},a_{13}}_{12~\text{meios}},a_{14})$

$a_1=-\dfrac{3}{4}$, $a_{14}=\dfrac{11}{6}$

$a_n=a_1+(n-1)r$

$\dfrac{11}{6}=-\dfrac{3}{4}+(14-1)r$

$13r=\dfrac{11}{6}+\dfrac{3}{4}$

$13r=\dfrac{22+9}{12}$

$r=\dfrac{31}{156}$

$a_1=-\dfrac{3}{4}$

$a_2=-\dfrac{3}{4}+\dfrac{31}{156}=-\dfrac{86}{156}$

$a_3=-\dfrac{86}{156}+\dfrac[31}{156}=-\dfrac{55}{156}$

$a_4=-\dfrac{55}{156}+\dfrac{31}{156}=-\dfrac{24}{156}$

$a_5=-\dfrac{24}{156}+\dfrac{31}{156}=\dfrac{7}{156}$

$a_6=\dfrac{38}{156}$

$a_7=\dfrac{69}{156}$

$a_8=\dfrac{100}{156}$

$a_9=\dfrac{131}{156}$

$a_{10}=\dfrac{162}{156}$

$a_{11}=\dfrac{193}{156}$

$a_{12}=\dfrac{224}{156}$

$a_{13}=\dfrac{255}{156}$

$a_{14}=\dfrac{286}{156}=\dfrac{11}{6}$