Linear recurrence sequences

Casio

Member
I have a linear recurrence sequence,

3, -1.5, 0.75, -375

x = a

a = 3

x2, = -1.5, x3, = 0.75, x4 = -375...

x2 = rx1+d
x3 = rx2+d

-1.5 = 3r + d
0.75 = -1.5 + d

-1.5 - 0.75 = (3r + d) - (-1.5 + d)

r = - 0.5

Sub in equation (2)

d = -1.5 - 3r = -1.5 - 3(-0.5)
d = 0

x4 = -0.5 x 0.75 + 0 = -0.375

Anyone see where I am going wrong!

P.S. I will practice the new form latex here as soon as I can.

Alexmahone

Active member
-1.5 = 3r + d
0.75 = -1.5 + d

Anyone see where I am going wrong!
Shouldn't that 2nd equation be 0.75 = -1.5r + d ?

Casio

Member
Shouldn't that 2nd equation be 0.75 = -1.5r + d ?
Thank you for that, typo error on my part, the result there = 0.75

But that is not helping me see where I am making a mistake?

Alexmahone

Active member
Thank you for that, typo error on my part, the result there = 0.75

But that is not helping me see where I am making a mistake?
So your sequence is a geometric progression with a=3 and r=-0.5. So I guess your answer of -0.375 is correct.

Last edited:

Casio

Member
So your sequence is a geometric progression with a=3 and r=-0.5. So I guess your answer of 0.375 is correct.
It seems my misunderstanding here is the answer = -0.375

I thought the answer should have been -375

These solutions are not the same?

soroban

Well-known member
Hello, Casio!

I have a linear recurrence sequence: .$3,\;\text{-}1.5,\;0.75,\;\text{-}375$ .??

There is NO WAY that the sequence is: .$3,\;-1\!\!\frac{1}{2},\;\frac{3}{4},\;\color{red}{-375}\;\cdots$

. . That last term must be -0.375 . . .

We have a geometric sequence with first term $a = 3$ and common ratio $r = -0.5$

The recurrence is: .$a_{n+1} \:=\:-0.5a_n$

Casio

Member
Hello, Casio!

There is NO WAY that the sequence is: .$3,\;-1\!\!\frac{1}{2},\;\frac{3}{4},\;\color{red}{-375}\;\cdots$

. . That last term must be -0.375 . . .

We have a geometric sequence with first term $a = 3$ and common ratio $r = -0.5$

The recurrence is: .$a_{n+1} \:=\:-0.5a_n$
Very sorry, you are right I copied the sequence incorrectly?

Ackbach

Indicium Physicus
Staff member
Very sorry, you are right I copied the sequence incorrectly?
Is that a statement or a question?