Question

What is the sum of a 6-term geometric series if the first term is 19 and the last term is 319,333? (3 points)
A) 372,552
B) 392,160
C) 411,768
D) 431,376
What is the sum of an 8-term geometric series if the first term is 15 and the last term is -4,199,040? (3 points)
A) -3,599,175
B) -3,359,230
C) -3,119,285
D) -2,879,340

Answer 1

find the common ratio

use \[T _{n}=ar^{n-1}\]

and the information for the 6th term

\[319333 = 19r^5\]

then

\[16807 = r^5\]

r = 7

so you know a = 19 r = 7 and n = 6

use the sum or a geometric series formula

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

substitute and evaluate for your answer

Answer 2

2235312 i got tht for an answer :/

Answer 3

well something is amiss...

Answer 4

are you sure there isn't a negative sign missing from 319333

Answer 5

nope

Answer 6

nope that doesn't work either

Answer 7

for B I found the common ratio to be -6

Answer 8

I got A for the first one.

Answer 9

And A for the second, also.

Answer 10

thank you so muchhh

Answer 11

for the 2nd question, a=15, n = 8 and r = -6 will give answer A

for question 1 I got answer A

\[s _{6}= \frac{19(7^6 -1)}{7-1}\]