sequences and series

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March 31st 2009, 03:56 AM
woollybull

sequences and series

Different numbers x, y and z are the first three terms of a geometric progression with common ration r, and also the first, second and fourth terms of an arithmetic progression.

a)Find the value of r.
b)Find which term of the arithmetic progression will next be equal to a term of the geometric progression.

a)
y=xr, z=xr^2

y=x+d,
z=x+3d
xr =x+d,
xr^2=x+3d
solve simulatenously for r=2.

b)
Is the only way to do this to assume that each geometric term is matched with an arithmetic term??
So I must find when x+(n-1)d=xr^3 (if r=2 then d=x)
Solving this for n= 8.

Please let me know if this is fine or if there is a better way.

Thanks.
March 31st 2009, 06:03 AM
HallsofIvy

Quote:

Originally Posted by woollybull

Different numbers x, y and z are the first three terms of a geometric progression with common ration r, and also the first, second and fourth terms of an arithmetic progression.

a)Find the value of r.
b)Find which term of the arithmetic progression will next be equal to a term of the geometric progression.

a)
y=xr, z=xr^2

y=x+d,
z=x+3d

xr =x+d,
xr^2=x+3d
solve simulatenously for r=2.

b)
Is the only way to do this to assume that each geometric term is matched with an arithmetic term??
So I must find when x+(n-1)d=xr^3 (if r=2 then d=x)
Solving this for n= 8.
Please let me know if this is fine or if there is a better way.

Thanks.

With r= 2, d= x, the geometric series is x, 2x, 4x, 8x, 16x, .... The arithmetic series is x, 2x, 3x, 4x, 5x, 7x, 8x, 9x, 10x. For this particular sequence, the arithmetic series will contain "nx" for all n. Yes, for this particular sequence, it is true that "each geometric term is matched by an arithmetic term". You could solve this without assuming that is true by writing out more terms as I have above and noting that "8x" is the next term that occurs in both sequences.
March 31st 2009, 06:23 AM
Soroban

Hello, woollybull!

Quote:

Different numbers $x, y, z$ are the first three terms of a G.P. with common ratio $r$ ,
and also the first, second and fourth terms of an A.P.

a) Find the value of $r.$

This is how I solved it . . .

. . $\begin{array}{|c||c|c|}\hline \text{Term} & \text{G.P.} & \text{A.P.} \\ \hline \hline x & a & a \\ \hline y & ar & a+d \\ \hline z & ar^2 & a+3d \\ \hline \end{array}$

We have: . $\begin{array}{ccccccccc}ar &=&a+d & \Rightarrow & d &=&a(r-1) & {\color{blue}[1]} \\ ar^2 &=& a+3d & \Rightarrow & 3d &=& a(r^2-1) & {\color{blue}[2]} \end{array}$

Divide [2] by [1]: . $\frac{3d}{d} \:=\:\frac{a(r^2-1)}{a(r-1)} \quad\Rightarrow \quad r^2-3r+2\:=\:0$

. . . . . . $(r-1)(r-2) \:=\:0 \quad\Rightarrow\quad r \:=\:1,\:2$

We are told that $x,y,z$ are different numbers, so: . $\boxed{r \:=\:2}$

And we find that: . $d \:=\:a$
The common difference is equal to the first term.

Quote:

b) Find which term of the A.P. will next be equal to a term of the G.P.
Do we assume that each geometric term is matched with an arithmetic term? . . . . no

The $m^{th}$ term of the G.P. is: . $a\!\cdot\!2^{m-1}$

The $n^{th}$ term of the A.P. is: . $a + (n-1)a \:=\:an$

They are equal when: . $a\!\cdot\!2^{m-1} \:=\:an \quad\Rightarrow\quad n \:=\:2^{m-1}$

If $m = 1$ , then $n = 1$ . . . This is the initial case.

When $m = 2$ , then $n = 2$ . . . $\boxed{\text{Their }second\text{ terms are equal.}}$

We can construct a table to display equal terms.

. . $\begin{array}{|c|c||c|}\hline \text{G.P.} & \text{A.P.} & \\ m & n & \text{Term}\\ \hline \hline 1 & 1 & a \\ 2 & 2 & 2a \\ 3 & 4 & 4a\\ 4 & 8 & 8a\\ 5 & 16 & 16a\\ \vdots & \vdots & \end{array}$