Question

Write an equation and determine the 8th term:
7, 7.5, 7.75, 7.875

Answer 1

All I really need to know is the common difference :s

Answer 2

The difference increases by a factor of 0.5.

Answer 3

hmmm.... 7.5 to 7.75 is just 0.25

Answer 4

But 7 to 7.5 is 0.5. :P

Answer 5

heheh =)

Answer 6

actually I noticed the ratio is an exponential

anyhow
\(\begin{array}{lllll}
&7+0.5^{1-1}&7+0.5^{2-1}&7+0.5^{3-1}&7+0.5^{4-1}\\
\hline\\
\implies &7& 7.5& 7.75& 7.875\\
\hline\\
n^{th}&1&2&3&4
\end{array}\)

Answer 7

There's no common difference.

But that pizza looks delicious.

@jdoe0001 it looks more like a series sum. 7+0.5+0.5^2+0.5^3...

Answer 8

well... hmm... well... sorta, yes

Answer 9

The next term is found by adding 0.5^(n-1)... Oh but it starts with 6,

\[\Large 6+0.5^0+0.5^1+0.5^2+0.5^3...\]

Answer 10

hmmm... ohh right... the 1st item.... will be 8 otherwise.

Answer 11

or one can say it starts with 7, and sequences out to the 2nd and forth

Answer 12

hmmm wait... nope you're right.. it starts with 6 either way

Answer 13

hmmm ahemm....

Answer 14

\(\bf \begin{array}{ccccccc}
&&\color{red}{(n-1)+0.5^{1-1}}&7+0.5^{2-1}&7.5+0.5^{3-1}&7.75+0.5^{4-1}\\
\hline\\
\implies &6&7& 7.5& 7.75& 7.875\\
\hline\\
n^{th}&1&2&3&4&5
\end{array}\)

Answer 15

woops.. forgot something

Answer 16

You can use a geometric sequence sum to find it, just add the 6 at the end... each term is just 6 plus the geometric sum to that point

ignoring the 6, the geometric sequence part is \[\Large a_n = a_1 r^{n-1}\]and the sum for a geometric is \[\Large a_1 \left( \frac{ 1-r^n }{ 1-r } \right)\]
your first term is 1, common ratio is 0.5.
\[\Large a_n = 0.5^{n-1}\]

Answer 17

he's offline right now, so no pizza for you yet =)

Answer 18

So the 8th term \[\Large 6+ \Large \left( \frac{ 1-0.5^8 }{ 1-8 } \right)\]

Answer 19

I guess a general formula would be \[\Large a_n = 6 + \Large \left( \frac{ 1-0.5^n }{ 1-0.5 } \right)\]

Answer 20

I think that works for all the sequence... never really needed to work out a sequence of this type tho, so the explanation isn't very clear as i was figuring it out as i went along.

Answer 21

hmm if I take the 7 as the 1st term.... \(\bf (n-1)+0.5^{n-1}\) seem to work out as the common ratio

Answer 22

Yeah i guess that would would too. But the above formula works and can be simplified to \[\Large a_n = 8-2*0.5^n \]

Answer 23

Looks better this way \[\Large a_n = 8-2(0.5^n)\]

Answer 24

@agent0smith @jdoe0001 oh my, seems complicated, but I think i got it down! thanks for the help :)