Question

I dont get it :(

Answer 1

So you are given this: \(\large a_1=38 \) and the formula \(\large a_n=\frac{1}{2}a_{n-1}\) when \(\large n \ge 2\).

If this was an arithmetic sequence, we would this formula: \(\large a_n= a_1+d(n-1) \)
but this is a *geometric* sequence, so we will use \(\large \color{red}{a_1*r^{n-1} }\) instead.

Plug in the first number, which is 2:
\(\large a_2= \frac{1}{2} a_1 \), and we were give that \(\large a_1=38\), so \(\large a_2=38*\frac{1}{2} =19\)

If you plug in 3:
\(\large a_3= \frac{1}{2} a_2 \), which we solved as \(\large a_2=19\) so it will be \(\large a_3=9.5 \)

If you continue, you will get \(\large a_4=4.75\), \(\large a_5=2.375\), etc..

Answer 2

But if you keep them in fraction form, you will notice a pattern:
\(\large a_2 = 38 \frac{1}{2} \)
\(\large a_3= 38 *\frac{1}{2} *\frac{1}{2} \)
\(\large a_4 = 38 *\frac{1}{2}*\frac{1}{2}*\frac{1}{2} \)
\(\large a_5 = 38 *\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}\)
and so on...
using \(\large \color{red}{a_n=a_1*r^{n-1}} \)

We can identity \(\large a_1\) as 38, and the r, which is what is being multiplied, as \(\large \frac{1}{2} \)

Giving:
\[\Large a_n=38*\frac{1}{2} ^{n-1} \]
when \(\large n \ge 2\)

Answer 3

Should be:

\[\Large a_n=38*(\frac{1}{2} )^{n-1}\]

Answer 4

its wrong btw

Answer 5

That's depressing, do you have the right one or..?

Answer 6

No clue all i know is that I hate math :)

Answer 7

the answer is...I do not know

Answer 8

I'm dumber than a nuT