Question

I have a small assignment to do for Algebra II. The topic is Geometric Series and I have a couple different questions... Please help!

Answer 1

For problems 1 – 4, solve. You must show your work to receive credit.

What is the sum of the geometric sequence 8, –16, 32 … if there are 15 terms? (1 point)
What is the sum of the geometric sequence 4, 12, 36 … if there are 9 terms? (1 point)
What is the sum of a 6-term geometric sequence if the first term is 11, the last term is –11,264 and the common ratio is –4? (1 point)
What is the sum of an 8-term geometric sequence if the first term is 10 and the last term is
781,250? (1 point)

Answer 2

@hartnn @nincompoop @campbell_st

Answer 3

well for a geometric series and the common ratio is greater then plus or minus 1 the sum is

\[S_{n} = \frac{a(1 - r^n)}{1 - r}\]

n = number of terms, a = 1st term and r = common ratio.

the common ratio is found by comparing terms

\[\frac{a_{2}}{a_{1}} = \frac{a_{3}}{a_{2}} ..... \]

or is each term being multiplied by the same value.

for the last question you'll need to find the common ratio using the formula for a term in a common ratio

\[a_{n} = a_{1}r^{n -1}\]

hope this helps

Answer 4

@campbell_st can you please walk me through one of them? I have the formula written down and I plugged in A and N, but how do I find the common ratio?

Answer 5

ok... so in the 1st question the 1st term is a = 8

what was 8 multiplied by to get -16..?

Answer 6

-2

Answer 7

yes ... and to check is the 3rd term -16 x -2 ...?

Answer 8

32

Answer 9

yes, it is

Answer 10

so you now have the common ratio, r = -2

substituting you get

\[S_{15} = \frac{8(1 - (-2)^{15})}{1 - (-2)}\]

all you need to do is take care with the negatives... and you'll get the sum

Answer 11

so it's S15=-87381.33?

Answer 12

wait... would it be 10,923?

Answer 13

well i got 87384