What is the partial sum for the geometric series in which a3=4, a4=-16, and n=6

Question

campbell_st · Answer

well do you know the common ratio..?

if not its 

$$\frac{a_{4}}{a_{3}} = $$

anonymous · Answer

-16/4=-4

campbell_st · Answer

ok thats great now you need to find the 1st term 

so 

the general term is 

$$a_{n} = a \times r^{n -1}$$

use the 3rd term... or the 4th term

so for the 3rd term = 4  and common ratio r = -4 and n = 3

you have

$$4 = a \times (-4)^{3 -1} $$

can you find the value of a from the equation above?

anonymous · Answer

a=1/4 ?

campbell_st · Answer

correct... all you need now is the partial sum, use the formula 

the formula is 

$$s_{n} = \frac{a(1 - r^n)}{1 - r}$$

you have a = 1/4, r = -4 and n = 6

substituting you get

$$s_{6} = \frac{\frac{1}{4}(1 - (-4)^6)}{1 - (-4)}$$

just evaluate

campbell_st · Answer

hope this helps...

anonymous · Answer

i got 204.85

campbell_st · Answer

well I got - 204.75

so me the sum should be negative since the 6th term is - 256

campbell_st · Answer

so you are adding 

1/4 + -1  + 4 + -16 + 64 + -256 = 

hope this helps.

anonymous · Answer

ok thank you!