geometric sequences (please help)

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November 6th 2008, 10:35 AM
juliak

geometric sequences (please help)

sequence 40, -20, 10. -5... is geometric
find Un and hence find the 10th term

what i got was
u1 = 40
and r = 1/2

so --> un=u1xr^n-1
un=40x(-1/2)^10-1
un=-20^9
un= -0.00000000000512
but the correct answer is -5/256
--> could you tell me where i got wrong?
November 6th 2008, 10:52 AM
Soroban

Hello, juliak!

I don't agree with their answer.

You were doing fine, then ...

Quote:

Geometric sequence: . $40,\:\text{-}20,\:10,\:\text{-}5,\:\hdots$

Find $U_n$ and hence find the $10^{th}$ term.

What i got was: . $U_1 = 40\:\text{ and }\:r = \text{-}\tfrac{1}{2}$ . . . . Good!

So: . $U_n \:=\:U_1r^{n-1}$

. . . $U_{10}\:=\:40\left(\text{-}\tfrac{1}{2}\right)^9$

We have: . $U_{10} \;=\;40\cdot\frac{1}{(\text{-}2)^9} \;=\;40\cdot\frac{1}{\text{-}512} \;=\;-\frac{40}{512} \;=\;-\frac{5}{64}$
November 6th 2008, 10:53 AM
Laurent

Quote:

Originally Posted by juliak

sequence 40, -20, 10, -5... is geometric
find Un and hence find the 10th term

what i got was
u1 = 40
and r = 1/2

Notice the change of sign in the sequence, hence $r=-\frac{1}{2}$ . But this is what you use then, so it must be a typo...

Quote:

so --> un=u1xr^n-1
un=40x(-1/2)^10-1
un=-20^9
un= -0.00000000000512
but the correct answer is -5/256
--> could you tell me where i got wrong?

This is a mistake in manipulating powers : $u_n=u_1 r^{n-1}=40\left(-\frac{1}{2}\right)^{10-1}=(-1)^9 \frac{40}{2^9}=-\frac{2^3\times 5}{2^9}=-\frac{5}{2^6}=-\frac{5}{64}$ (I use $40=8\times 5=2^3\times 5$ to simplify the ratio).
November 6th 2008, 11:07 AM
juliak

^ how did you get from
40(-1/2)^9 to (-1)^9(40/2^9)
?
November 6th 2008, 11:32 AM
Laurent

Quote:

Originally Posted by juliak

^ how did you get from
40(-1/2)^9 to (-1)^9(40/2^9)
?

The power of a product is the product of the powers: $(ab)^n=a^nb^n$ . And the same for ratios: $\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$ . Here, we have $40\left(-\frac{1}{2}\right)^9=40(-1)^9\left(\frac{1}{2}\right)^9=40(-1)^9\frac{1}{2^9}$ . Ok?