Question

Identify the 16th term of a geometric sequence where a1 = 4 and a8 = −8,748.

−172,186,884
−57,395,628
57,395,628
172,186,884

Answer 1

You still post questions here? smh

Answer 2

@TheInspector yep cx

Answer 3

@johnweldon1993

Answer 4

Any ideas?

Knowing the formula we want to use is \(\large a_n = a_1 * r^{n - 1}\)

Answer 5

4 * ??

Answer 6

\[a^8 / a^1 = ar^7 / a = r^7\]

Answer 7

Youʻre on the right track

Answer 8

First we need to solve for the 'r' *common ratio*

So plug in what we know

\[\large a_8 = a_1 * r^{8-1}\]
\[\large -8748 = 4r^{7}\]

Since we knew what a8 and a1 were...what would solving that for 'r' result in?

Answer 9

;c idk. im so lost.

Answer 10

Common Ratio: 7th root of

\[\frac{ -8748 }{ 4 } = -3\]

Answer 11

Alright, so starting from the beginning...to find the n'th term in a geometric sequence...you use the formula \(\large a_n = a_1 * r^{n-1}\)

This says...to find the n'th term...take the first term in the sequence and multiply it by the common ratio raised to the n'th - 1 power

So...we are given \(\large a_8 = -8748\) and \(\large a_1 = 4\)

So we need to solve for 'r' first

Plug in what we know

\[\large -8748 = 4r^{8-1}\]
which becomes
\[\large -8748 = 4r^7\]

Or

\[\large r = \sqrt[7]{-8748}\]

What does r equal?

Answer 12

oops...my mistake up there...i never divided by the 4!

Answer 13

\[\large r = \sqrt[7]{\frac{-8748}{4}}\]

Answer 14

haha, yeah

Answer 15

idk how to solve for r tho..

Answer 16

-8748 = 4r^7

i understand how we got this.. now what

Answer 17

Okay so from that step...divide both sides by 4

\[\large -2187 = r^7\]

So to solve for 'r'...we take the 7th root of that number

Answer 18

Same thing as taking a square root basically

we know \(\large \sqrt[2]{4} = 2\)

This time it's just a bigger number...but just plug it into your calculator :)

Answer 19

Or google of course :P and you'll see it comes out to -3

Answer 20

ok so r is -3

Answer 21

-3 * 4 = -12

-12^7

= -35,831,808

Answer 22

Right

So now go back to the beginning

\[\large a_n = a_1*r^{n-1}\]

We now have a_1, we just found 'r'...we have all we need to solve for \(\large a_16\)

\[\large a_{16} = a_1 * r^{16 - 1}\]
\[\large a_{16} = 4*(-3)^{15}\]

Answer 23

You were still thinking about that 8th term we used in the beginning...that's trash now we dont need it now that we found 'r' :P

Just remember...for these problems we need \(\large a_1\) and we need \(\large r\) and that's it

Now that we have them...disregard everything else and only use those to find what you need

Answer 24

ok answer is B.. thank you..