Question

Geometric sequence question, plz answer

Answer 1

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

Answer 2

a10 = 1024/729 a10 =- 1024/729

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