Question

Find the 6th term of a geometric sequence t3 = 444 and t7 = 7104

Answer 1

will give medal

Answer 2

The quotient of terms t7 and t3 yields 16. 7-3=4. Since \[2^4=16\], we know we are doubling the result to achieve the next term. Thus, working backwards from t7, we can divide it by 2 to yield t6=3552. OR, working forward from t3, we can double it 3 times to achieve the same result, t6=3552.

Answer 3

In other words, we started this series with t1=111. t2=222, t3=444, t4=888, t5=1776, t6=3552, and t7=7104.

Answer 4

Find the 6th term of the sequence in which t1 = -5 and tn = -5tn-1

Answer 5

and for the first question it says in my directions there are two possible ways that i need to list

Answer 6

t1=-5, t2=-5(-5)=25, t3=-5(25)=-125, t4=-5(125)=625, t5=-5(625)=-3125, t6=-5(-3125)=15625. In other words, a closer form formula for \[t_{n}=(-5)^n\]

Answer 7

That is to say, a CLOSED-form formula is the above. Thus, \[t_{6}=(-5)^6=15625\]

Answer 8

im confused

Answer 9

On which part, the first problem or the second?

Answer 10

the first.. it asks in the question for two possible ways and you only showed me one...

Answer 11

A geometric sequence takes the form \[a_{n}=a_{0}r^n\]Thus, each subsequent term has been multiplied by r. Using that formula and the information you were provided (specifically, that \[t_{3}=444\]and \[t_{7}=7104\]we could then say \[t_{3}=444=t_{0}r^3\]and\[t_{7}=7104=t_{0}r^7\]Now, we don't know what \[t_{1}\] is specifically, however, using the above statements we could write it as \[t_{1}=\frac{ 444 }{ r^3 }\]and\[t_{1}=\frac{ 7104 }{ r^7 }\]Equating these two statements, we have \[\frac{ 444 }{ r^3 }=\frac{ 7104 }{ r^7 }\]which is equivalent to saying \[444r^7=7104r^3\]Dividing both sides by \[r^3\](which we can do because we're going to assume \[r \neq0\]as otherwise this wouldn't be a geometric series), thus resulting in \[444r^4=7104\]Finally, divide by 444 and get \[r^4=16\]Thus, \[r=2\]Now, using the general formula for a geometric sequence, we know the following about this sequence:\[t_{n}=t_{0}2^n\]Given that \[t_{3}=444=t_{1}2^3=8t_{0}\]Thus, our starting (or "seed") value is \[t_{0}=\frac{ 111 }{ 2 }\]Therefore, we now have all parts of our generic equation, and plugging these in we get a closed-form formula:\[t_{n}=\frac{ 111 }{ 2 }2^n\]This gives us\[t_{6}=3552\]

Answer 12

I'm sorry, that was to say, "we don't know what \[t_{0}\] is specifically, although we know that \[t_{0}=\frac{ 444 }{ r^3 }\]and\[t_{0}=\frac{ 7104 }{ r^7 }\]

Answer 13

I know this second way is more difficult, but it utilizes the definition of a geometric sequence, so it's more mathematically correct. The bottom line is that adjacent terms of a geometric sequence vary only by a factor of r, whatever that happens to be (in this case, r=2). Therefore, if you have terms that are TWO apart (let's say t3 and t5 for example), they will have a ratio of r squared (in this case, r^2=4). My first explanation took advantage of this fact. I found the ratio of t7 and t3 (which was 16). I then noted that they are 4 places apart in the sequence. \[\log_{2}(16)=4 \]So it's not a coincidence that they are 4 places apart in the sequence. Thus, I can move backward or forward in the sequence by dividing or multiplying by 2, respectively.