(11C) n-th term of a Geometric Sequence
05-21-2018, 09:31 AM (This post was last modified: 05-22-2018 11:52 AM by Gamo.)
Post: #1 Gamo Senior Member Posts: 720 Joined: Dec 2016
(11C) n-th term of a Geometric Sequence
Nth term of a Geometric Sequence giving
Initial Value (a)
Common Ratio (r)
Term (n)

Formula: a sub n = ar^n-1 for every integer n ≥ 1

https://en.wikipedia.org/wiki/Geometric_progression

Example:
a=2
r=3.14
n=14

Procedure:
Set User Mode
2 A > 2 // Input initial value
3.14 B > 3.14 // Input common ratio
14 C > 14 // Input term
D > 5769197.69 // answer on the 14th term

What is the sequence number on the 1st and 2nd term on the above example?

(1st Term) 1 C > D > 2
(2nd Term) 2 C > D > 6.28

The sequence are 2, 6.28, 19.72, 61.92,.......

Program:
Code:
 LBL A  // Initial Value STO 1 RTN LBL B  // Common Ratio STO 2 RTN LBL C  // n-term STO 3 RTN LBL D RCL 2 X>0?   // Test Common Ratio for Positive or Negative value  GTO 1 RCL 3 RCL 3 2 ÷ FRAC 2 x X=0?    // Test n-th term for Even or Odd number GTO 3 GSB 2    // Odd integer of n-th term RCL 1 x RTN LBL 1    // Common Ratio is Greater than Zero RCL 2 RCL 3 1 - Y^X RCL 1 x RTN LBL 2    // identical routine for common ratio value Less than Zero RCL 2 RCL 3 1 - X<>Y CHS X<>Y Y^X RTN LBL 3    // Even integer of n-th term GSB 2 CHS RCL 1 x RTN

Remark:
The common ratio of a geometric sequence may be negative, resulting in an alternating sequence, with numbers switching from positive to negative and back. For instance

1, −3, 9, −27, 81, −243, ...
is a geometric sequence with common ratio −3.

Input 1 A, -3 B, and your choice of term [C] then [D] for answer.

------------------------------------------------------------------------------
This update version can be use for the following:

LBL A // To display a geometric progression
LBL B // To find from the n-th term
LBL C // To find the sum of the giving n-th term

Procedure:
For geometric sequence: (a) ENTER (r) > f [A] > continue [R/S]
For n-th term: (a) ENTER (r) ENTER (n) > f [B]
For the Sum of given n-th term: (a) ENTER (r) ENTER (n) > f [C]

Example:
a=1
r= -3
n= selected term

Sequence: 1, -3, 9, -27, 81, -243,......

Geometric Sequence: 1 ENTER -3 f [A] > 1 > [R/S] > -3 > [R/S] > 9
4th term: 1 ENTER -3 ENTER 4 f [B] > -27
Sum to 4th term: 1 ENTER -3 ENTER 4 f [C] > -20

Program:
Code:
 LBL A  // Geometric Sequence  ENTER ENTER R^ LBL 1 R/S x GTO 1 LBL B    // n-th term Rv Rv STO 3 Rv Rv 1 - GSB 2 RCL 3 x RTN LBL C    // Sum to the n-th term Rv STO 4 Rv STO 3 Rv Rv GSB 2 1 - RCL 3 x RCL 4 1 - ÷ RTN LBL 2 STO 1 X<>Y STO 2 0 X<>Y X>Y GTO 3 CHS STO 2 RCL 1 2 ÷ FRAC 0 X=Y GTO 3 RCL 2 RCL 1 Y^X CHS RTN LBL 3 RCL 2 RCL 1 Y^X RTN

Gamo
05-21-2018, 12:47 PM (This post was last modified: 05-21-2018 01:31 PM by Dieter.)
Post: #2
 Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: (11C) n-th term of a Geometric Sequence
(05-21-2018 09:31 AM)Gamo Wrote:  Formula: a sub n = ar^n-1 for every integer n ≥ 1

an = a1 · rn–1

(05-21-2018 09:31 AM)Gamo Wrote:
Code:
LBL D RCL 2 X>0?   // Test Common Ratio for Positive or Negative value  GTO 1 ... X=0?    // Test n-th term for Even or Odd number GTO 3 GSB 2    // Odd integer of n-th term RCL 1 x RTN LBL 1    // Common Ratio is Greater than Zero ... LBL 2    // identical routine for common ratio value Less than Zero ... LBL 3    // Even integer of n-th term GSB 2 CHS RCL 1 x RTN

Gamo, all these tests and subroutines are not required. You do not have to check whether r is negative or not, and if it is, also check if n is odd or even. All HP calculators I know can calculate integer powers of negative bases directly. So (–3,14)5 directly yields –305,24... without any problem.

This means the more than 40 steps routine at LBL D can be replaced with this:

Code:
LBL D RCL 2 RCL 3 1 - Y^X RCL 1 x RTN

That's it.

If you really want to check for odd or even n and whether r is negative or not, you could do it much shorter. Simply calculate the power with |r| and change the sign if r<0 and n is even. For instance this way:

Code:
LBL D CF 0 RCL 2 X>0? GTO 1  // keep flag 0 clear for r>0 RCL 3 2 / FRAC   // for r<0 check if n is odd or even X=0?   // if n is even SF 0   // set flag 0 LBL 1 RCL 2 ABS    // calculate |r|^(n-1) RCL 3 1 - Y^X RCL 1 x F? 0 CHS    // change sign if r<0 and n even CF 0 RTN

But again: all this is not required.

You know I'm a big fan of using the stack instead of registers, so here's how I would do it:

Code:
LBL A ENTER 1 - X<>Y ENTER R↓ X<>Y Y^X x RTN

The ENTER in the second line usually is not required, but this way the program will also run on an HP25. ;-)

Usage:
a1 [ENTER] r [ENTER] n
[A] => an
[x] => an+1
[x] => an+2
...

Example:
1 [ENTER] –3 [ENTER] 1
[A]   1
[x] –3
[x]   9
[x] –27
[x]   81
[x] –243
...

Dieter
05-21-2018, 03:42 PM
Post: #3 Gamo Senior Member Posts: 720 Joined: Dec 2016
RE: (11C) n-th term of a Geometric Sequence
Dieter Thank You

I like the shorter solution for even or odd routine.

Your version of the step through a geometric progression is good but my manual version is easier.

Steps:
(r) ENTER ENTER ENTER
(a)
x (multiply) keep repeat this x as desire

Example:
1, −3, 9, −27, 81, −243, ...

-3 ENTER ENTER ENTER
1
X (repeat multiplication)

Gamo
05-21-2018, 04:06 PM (This post was last modified: 05-21-2018 04:08 PM by Dieter.)
Post: #4
 Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: (11C) n-th term of a Geometric Sequence
(05-21-2018 03:42 PM)Gamo Wrote:  I like the shorter solution for even or odd routine.

Maybe it has not become clear yet: You do not need this odd/even thing at all!
The 11 step program does it all.

(05-21-2018 03:42 PM)Gamo Wrote:  Your version of the step through a geometric progression is good but my manual version is easier.

Steps:
(r) ENTER ENTER ENTER
(a)
x (multiply) keep repeat this x as desire

Sure... if you start at the beginning of the sequence, i.e. n=1. With the program you can start at, say, n=20 and continue from there.

Dieter
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