This is the first round for series of posts about optimizing the use of calculator in solving math problems. The calculator techniques I am presenting here has been known to many students who are about to take the engineering board exam. Using it will save you plenty of time and use that time in analyzing more complex problems. The following models of CASIO calculator may work with these methods: fx570ES, fx570ES Plus, fx115ES, fx115ES Plus, fx991ES, and fx991ES Plus.
This post will focus on progression progression. To illustrate the use of calculator, we will have sample problems to solve. But before that, note the following calculator keys and the corresponding operation:
Name 
Key 
Operation 

Shift 

SHIFT 
Mode 

MODE 
Alpha 

ALPHA 
Stat 

SHIFT → 1[STAT] 
AC 

AC 

Name 
Key 
Operation 

Σ (Sigma) 

SHIFT → log 
Solve 

SHIFT → CALC 
Logical equals 

ALPHA → CALC 
Exponent 

x^{[]} 

Problem: Arithmetic Progression
The 6^{th} term of an arithmetic progression is 12 and the 30^{th} term is 180.
1. What is the common difference of the sequence?
2. Determine the first term?
3. Find the 52^{nd} term.
4. If the n^{th} term is 250, find n.
5. Calculate the sum of the first 60 terms.
6. Compute for the sum between 12th and 37th terms, inclusive.
Traditional Solution
For a little background about Arithmetic Progression, the traditional way of solving this problem is presented here.
Calculator Technique for Arithmetic Progression
Among the many STAT
type, why A+BX
?
The formula a_{n} = a_{m} + (n  m)d is linear in n. In calculator, we input n at X column and a_{n} at Y column. Thus our X is linear representing the variable n in the formula.
Bring your calculator to Linear Regression in STAT mode:
MODE → 3:STAT → 2:A+BX
and input the coordinates.
X (for n) 
Y (for a_{n}) 
6 
12 
30 
180 
To find the first term:
AC → 1 SHIFT → 1[STAT] → 7:Reg → 5:ycaret
and calculate 1ycaret
, be sure to place 1 in front of ycaret.
1ycaret
= 23 → answer for the first term
To find the 52^{nd} term, and again AC → 52 SHIFT → 1[STAT] → 7:Reg → 5:ycaret
and make sure you place 52 in front of ycaret.
52ycaret
= 334 → answer for the 52^{nd} term
To find n for a_{n} = 250,
AC → 250 SHIFT → 1[STAT] → 7:Reg → 4:xcaret
250xcaret
= 40 → answer for n
To find the common difference, solve for any term adjacent to a given term, say 7th term because the 6th term is given then do 7ycaret
 12 = 7 for d. For some fun, randomly subtract any two adjacent terms like 18ycaret
 17ycaret
, etc. Try it!
Sum of Arithmetic Progression by Calculator
Sum of the first 60 terms:
AC → SHIFT → log[Σ] → ALPHA → )[X] → SHIFT → 1[STAT] → 7:Reg → 5:ycaret → SHIFT → )[,] → 1 → SHIFT → )[,] → 60 → )
The calculator will display Σ(Xycaret,1,60)
then press [=]
.
Σ(Xycaret,1,60) =
11010 ← answer
Sum from 12th to 37th terms,
Σ(Xycaret,12,37) =
3679 ← answer
Another way to solve for the sum is to use the Σ calculation outside the STAT mode. The concept is to add each term in the progression. Any term in the progression is given by a_{n} = a_{1} + (n  1)d. In this problem, a_{1} = 23 and d = 7, thus, our equation for a_{n} is a_{n} = 23 + (n  1)(7).
Reset your calculator into general calculation mode: MODE → 1:COMP
then SHIFT → log
.
Sum of first 60 terms:
$\displaystyle \sum_{x=1}^{60}$ (23 + (ALPHA X  1) × 7)
= 11010
Or you can do
$\displaystyle \sum_{x=0}^{59}$ (23 + 7 ALPHA X)
= 11010 which yield the same result.
Sum from 12th to 37th terms
$\displaystyle \sum_{x={12}}^{37}$ (23 + (ALPHA X  1) × 7)
= 3679
Or you may do
$\displaystyle \sum_{x=11}^{36}$ (23 + 7 ALPHA X)
= 3679
Calculator Technique for Geometric Progression
Problem
Given the sequence 2, 6, 18, 54, ...
1. Find the 12th term
2. Find n if a_{n} = 9,565,938.
3. Find the sum of the first ten terms.
Traditional Solution
Solution by Calculator
Why A·B^X?
The nth term formula a_{n} = a_{1}r^{n – 1} for geometric progression is exponential in form, the variable n in the formula is the X equivalent in the calculator.
MODE → 3:STAT → 6:A·B^X
To solve for the 12^{th} term
AC → 12 SHIFT → 1[STAT] → 7:Reg → 5:ycaret
12ycaret
= 354294 answer
To solve for n,
AC → 9565938 SHIFT → 1[STAT] → 7:Reg → 4:xcaret
9565938xcaret
= 15 answer
Sum of the first ten terms,
AC → SHIFT → log[Σ] → ALPHA → )[X] → SHIFT → 1[STAT] → 7:Reg → 5:ycaret → SHIFT → )[,] → 1 → SHIFT → )[,] → 10 → )
The calculator will display Σ(Xycaret,1,10)
then press [=]
.
Σ(Xycaret,1,10) =
59048 ← answer
You may also sove the sum outside the STAT mode
(MODE → 1:COMP
then SHIFT → log[Σ]
)
Each term which is given by a_{n} = a_{1}r^{n – 1}.
$\displaystyle \sum_{x=1}^{10}$(2(3^{ALPHA X  1}))
= 59048 answer
Or you may do
$\displaystyle \sum_{x=0}^{9}$(2 × 3^{ALPHA X})
= 59048
Calculator Technique for Harmonic Progression
Problem
Find the 30th term of the sequence 6, 3, 2, ...
Solution by Calculator
MODE → 3:STAT → 8:1/X
AC → 30 SHIFT → 1[STAT] → 7:Reg → 5:ycaret
30ycaret
= 0.2 answer
I hope you find this post helpful. With some practice, you will get familiar with your calculator and the methods we present here. I encourage you to do some practice, once you grasp it, you can easily solve basic problems in progression.
If you have another way of using your calculator for solving progression problems, please share it to us. We will be happy to have variety of ways posted here. You can use the comment form below to do it.
Comments
Re: Calculator Techniques for Solving Progression Problems
thank you... very helpful!
Re: Calculator Techniques for Solving Progression Problems
I have a problem in geometric progression, i got a math error because the value of my r is negative. Here is the example
Find the sum of the first 7 terms of the sequence 9, 6, 4,...
Mode 3 6 is not applicable
Mode 3 6 is not applicable for negative common ratio. You can use it though but you need to exclude the sign and get your actual sign manually. This approach however is not applicable for for finding the sum of GP.
thank you much
thank you much
thank you very much sir!
thank you very much sir! sobrang laking tulong po. Godbless!
Sir, how about the technique
Sir, how about the technique when the given is SUM or when the problem requires to find the sum of odd numbers of the series?