Respuesta :

wegnerkolmp2741o

Answer:

a10 = 1024/729                                   a10 =- 1024/729

Step-by-step explanation:

Formula for geometric sequence

an = a r^n

The third term is

24 = a r^3

The fifth terms is

32/3  =a r^5

Divide these equations

32/3 = ar^5

--------------------

24  = a r^3

32/3

------  = r^2

24

Copy dot flip

32/3 * 1/24  = r^2

Divide the top and bottom by 8

4/3*3 = r^2

2*2/(3*3) = r^2

Take the square root of each side

sqrt (2*2/(3*3) =sqrt (r^2)

± 2/3 = r

Now we need to solve for a

an = a (2/3)^n     or                      an = a (-2/3)^n

Using the third term

24 = a (2/3)^3                          24 = a (-2/3)^3  

24 = a (8/27)                              24 = a (-8/27)

Multiply by 27/8                      Multiply by -27/8

24*27/8 = a                             24*-27/8 = a

81=a                                                     -81=a

an = 81 (2/3)^n     or                      an = -81 (-2/3)^n

Now we want to find the 10th term.  Let n =10

a10 = 81 (2/3)^10     or                      a10 = -81 (-2/3)^10

a10 = 1024/729                                   a10 =- 1024/729

gmany

[tex]t_n=t_1r^{n-1}\\\\\text{We have}\ t_3=24\ \text{and}\ t_5=\dfrac{32}{3}.\\\\\dfrac{t_5}{t_3}=r^2\\\\\text{Substitute:}\\\\r^2=\dfrac{\frac{32}{3}}{24}\\\\r^2=\dfrac{32}{3}\cdot\dfrac{1}{24}\\\\r^2=\dfrac{4}{3}\cdot\dfrac{1}{3}\\\\r^2=\dfrac{4}{9}\to r=\pm\sqrt{\dfrac{4}{9}}\to\ r=-\dfrac{2}{3} \ or\ r=\dfrac{2}{3}\\\\t_{10}=t_5r^5\\\\\text{substitute}\\\\t_{10}=\dfrac{32}{3}\cdot\left(\pm\dfrac{2}{3}\right)^5=\dfrac{32}{3}\cdot\left(\pm\dfrac{32}{243}\right)=\pm\dfrac{1024}{729}\\\\Answer:\ \boxed{t_{10}=-\dfrac{1024}{729}\ or\ \dfrac{1024}{729}}[/tex]

Otras preguntas