Question

In a geometric sequence, a(down)7=14 and r = – 1/2. What is the 10th term of the sequence?

Answer 1

\[\LARGE a_7=14 \]and
\[\LARGE r=\frac{-1}{2}\]
for \[\LARGE a_{10}=ar^{9}\]
can you find a from given conditions?
\[\LARGE a_7=ar^{n-1}\]

Answer 2

you are given a7=14
and r=-1/2 and n=6

Answer 3

yes

Answer 4

a=?

Answer 5

I'm confused on how to fill it out correctly

Answer 6

\[\LARGE 14=a(\frac{1}{2})^6\]

Answer 7

calculate a

Answer 8

ok 1 sec

Answer 9

its wrong,sorry

Answer 10

\[\LARGE a=14 \times (2)^6\]

Answer 11

relax urself :) take your time!

Answer 12

I'm confused

Answer 13

yes :)

Answer 14

a=896

Answer 15

now substitute here..
\[\LARGE a_{10}=ar^{9}\]
a=896
r=-1/2

Answer 16

I keep getting the wrong answer :(

Answer 17

what did you get?

Answer 18

\[x_{10}=-\frac{ 1 }{ 512 }\]

Answer 19

?????how

Answer 20

can you show me your work

Answer 21

\[\LARGE a_{10}=896 \times \frac{-1}{512}\]

Answer 22

so it's b. -1.75

Answer 23

The geometric sequence is
\[a_n = ar^{n-1}\]

Given:
\[r = -\frac{1}{2}\]
\[a_7 = ar^6 =14\]
Substituting r into ar^6 gives:
\[a\left(-\frac{1}{2}\right)^6 = 14\]
\[a = 14 * 2^6 = 14 * 64 = 896\]

Substiting a and r into the definition of a geometric sequence gives us the sequence we're after:
\[a_n = 896\left(-\frac{1}{2}\right)^{n-1}\]

so
\[a_{10} = 896\left(-\frac{1}{2}\right)^9 = -\frac{896}{512} = -1\frac{3}{4} \]