How many terms are in the geometric sequence having a first term of 2, a last term of 32, and a common ratio of -2?

3

Step-by-step explanation:

first term a= 2

last term tn= 32

common ratio r= -2

number of terms

= tn=ar^n-1

32 = 2*(-2)^n-1

divide both sides by 2

32/2 = 2*(-2)^n-1/2

16 = (-2)^n-1

Then...use 16 in the power of 2

(-2)^4 = (-2)^n-1

The bases (-2), cancels out

so we have: 4 = n-1

collect like terms

4-1 = n

3 = n

n= 3

i am currently in the exact same situation rn. algebra2/adv algebra is killing

step-by-step explanation:

hey there!

if she's saving for six months, let's divide the total number we need for the trip by six, dividing it into the amount that needs to be saved for each month.

500 ÷ 6 =

she needs to save at least this amount each month, so we should use this symbol: ≥

let x be the amount she saves each month.

your final answer is x ≥

hope this !