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: fx-570ES, fx-570ES Plus, fx-115ES, fx-115ES Plus, fx-991ES, and fx-991ES 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<sup>[]</sup>`

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.

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$a_n = a_m + (n - m)d$

$a_{30} = a_6 + (30 - 6)d$

$180 = 12 + 24d$

$d = 7$ → common difference

$a_n = a_1 + (n - 1)d$

$a_{30} = a_1 + (30 - 1)d$

$180 = a_1 + 29(7)$

$a_1 = -23$ → first term

$a_n = a_m + (n - m)d$

$a_{52} = a_6 + (52 - 6)d$

$a_{52} = 12 + 46(7)$

$a_52 = 334$ → 52^nd term

$a_n = a_m + (n - m)d$

$250 = 12 + (n - 6)(7)$

$n = 40$ → 40^th term, a₄₀ = 250

Sum of AP is given by the formula
$S = \frac{1}{2}n[ \, 2a_1 + (n - 1)d \, ]$

Sum of the first 60 terms
$S = \frac{1}{2}(60)[ \, 2(-23) + (60 - 1)(7) \, ]$

$S = 11,010$ → answer

Sum between 12th and 37th terms, inclusive.
$S = S_{37} - S_{11}$

$S = \frac{1}{2}(37)[ \, 2(-23) + (37 - 1)(7) \, ] - \frac{1}{2}(11)[ \, 2(-23) + (11 - 1)(7) \, ]$

$S = 3679$ → answer

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:y-caret and calculate 1y-caret, be sure to place 1 in front of y-caret.

1y-caret = -23 → answer for the first term

To find the 52^nd term, and again AC → 52 SHIFT → 1[STAT] → 7:Reg → 5:y-caret and make sure you place 52 in front of y-caret.

52y-caret = 334 → answer for the 52^nd term

To find n for a_n = 250,
AC → 250 SHIFT → 1[STAT] → 7:Reg → 4:x-caret

250x-caret = 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 7y-caret - 12 = 7 for d. For some fun, randomly subtract any two adjacent terms like 18y-caret - 17y-caret, 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:y-caret → SHIFT → )[,] → 1 → SHIFT → )[,] → 60 → )

The calculator will display Σ(Xy-caret,1,60) then press [=].
Σ(Xy-caret,1,60) = 11010 ← answer

Sum from 12th to 37th terms,
Σ(Xy-caret,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 = a₁ + (n - 1)d. In this problem, a₁ = -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

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Common ratio, $r = 6/2 = 3$

$a_n = a_1r^{n - 1}$

$a_{12} = a_1r^{12 - 1}$

$a_{12} = (2)(3^11)$

$a_{12} = 354,294$ answer

$a_n = a_1r^{n - 1}$

$9\,565\,938 = 2(3^{n - 1})$ → Line (2)

$4\,782\,969 = 3^{n - 1}$

$\ln 4\,782\,969 = \ln 3^{n - 1}$

$\ln 4\,782\,969 = (n - 1)\ln 3$

$n = \dfrac{\ln 4\,782\,969}{\ln 3} + 1$

$n = 15$ answer

Or you may use SHIFT → SOLVE of the calculator in Line 2 to find n directly

9565938 ALPHA CALC 2(3ALPHA X-1) SHIFT → SOLVE =

wait for a little while... Calculator will output

X = 15
L - R = 0

Answer for n is 15

$S = \dfrac{a_1(r^n - 1)}{r - 1}$

$S = \dfrac{2(3^{10} - 1)}{3 - 1}$

$S = 59\,048$ answer

Solution by Calculator

Why A·B^X?
The nth term formula a_n = a₁r^{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

X	Y
1	2
2	6
3	18

To solve for the 12^th term
AC → 12 SHIFT → 1[STAT] → 7:Reg → 5:y-caret
12y-caret = 354294 answer

To solve for n,
AC → 9565938 SHIFT → 1[STAT] → 7:Reg → 4:x-caret
9565938x-caret = 15 answer

Sum of the first ten terms,
AC → SHIFT → log[Σ] → ALPHA → )[X] → SHIFT → 1[STAT] → 7:Reg → 5:y-caret → SHIFT → )[,] → 1 → SHIFT → )[,] → 10 → )

The calculator will display Σ(Xy-caret,1,10) then press [=].

Σ(Xy-caret,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 = a₁r^{n – 1}.

$\displaystyle \sum_{x=1}^{10}$(2(3ALPHA X - 1)) = 59048 answer

Or you may do
$\displaystyle \sum_{x=0}^{9}$(2 × 3ALPHA 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

X	Y
1	6
2	3
3	2

AC → 30 SHIFT → 1[STAT] → 7:Reg → 5:y-caret

30y-caret = 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.

Calculator Techniques for Solving Progression Problems

Calculator Technique for Arithmetic Progression

Calculator Technique for Geometric Progression

Calculator Technique for Harmonic Progression

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Calculator Technique for Arithmetic Progression

Calculator Technique for Geometric Progression

Calculator Technique for Harmonic Progression

Re: Calculator Techniques for Solving Progression Problems

Mode 3 6 is not applicable

thank you much

thank you very much sir!

Sir, how about the technique

In harmonic Progression, do I

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