Home Contact Glossary Blogs Popular Recent About
Calculator Techniques for Solving Progression
Problems
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[]
Problem: Arithmetic Progression
The 6th term of an arithmetic progression is 12 and the 30th term is 180. 1. What is the common difference of the sequence?
2. Determine the first term? 3. Find the 52nd term. 4. If the nth 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.
Algebra
www.pubget.com/PaperStore
Order PDFs Papers Delivered in Minutes
Click here to show or hide the solution
Among the many STAT type, why A+BX? The formula an = am + (n - m)d is linear in n. In
calculator, we input n at X column and an at Y column.
Thus our X is linear representing the variable n in the formula.
Why MODE → 3:STAT → 3:_+cX2?
The formula S = ½n[ 2a1 + (n - 1)d ] for sum of
arithmetic progression is quadratic in n. In our calculator, we input n in the X column and the sum at the Y column.
Calculator Technique for Arithmetic Progression
Bring your calculator to Linear Regression in STAT mode:MODE → 3:STAT → 2:A+BX and input the coordinates. X (for n) Y (for an)
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 52nd 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 52nd term
To find n for an = 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
Bring the your calculator to Quadratic Regression in STAT mode
MODE → 3:STAT → 3:_+cX2
Note that for the given AP, a1 = -23, a2 = -16, and a3 = -9. Input three
coordinates
X Y
1 -23
2 -23-16
3 -23-16-9
Sum of the first 60 terms: (AC → 60 SHIFT → 1[STAT] → 7:Reg → 6:y-caret) 60y-caret = 11010
→ first term
→ 52nd term
→ 40th term, a40 = 250
Sum of AP is given by the formula
Sum of the first 60 terms → answer
Sum between 12th and 37th terms, inclusive.
Why A·B^X?
The nth term formula an = a1rn – 1 for geometric
progression is exponential in form, the variable n in the formula is the X equivalent in the calculator.
Sum from 12th to 37th terms, use SHIFT → 1[STAT] → 7:Reg → 6:y-caret twice
37y-caret - 11y-caret = 3679
Another way to solve for the sum is to use the Σ calculation. The concept is to add each term in the progression. Any term in the progression is given by an = a1 +
(n - 1)d. In this problem, a1 = -23 and d = 7, thus, our equation for an is an = -23 + (n - 1)(7).
Reset your calculator into general calculation mode: MODE → 1:COMP then SHIFT → log. Sum of first 60 terms:
(-23 + (ALPHA X - 1) × 7) = 11010
Or you can do
(-23 + 7 ALPHA X) = 11010 which yield the same result.
Sum from 12th to 37th terms
(-23 + (ALPHA X - 1) × 7) = 3679
Or you may do
(-23 + 7 ALPHA X) = 3679
Calculator Technique for Geometric Progression
ProblemGiven the sequence 2, 6, 18, 54, ... 1. Find the 12th term
2. Find n if an = 9,565,938.
3. Find the sum of the first ten terms. Traditional Solution
Solution by Calculator
MODE → 3:STAT → 6:A·B^X
X Y
1 2
2 6
3 18
To solve for the 12th 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 (MODE → 1:COMP then SHIFT → log) Each term which is given by an = a1rn – 1.
(2(3ALPHA X - 1)) = 59048 answer
Or you may do
(2 × 3ALPHA X) = 59048
Calculator Technique for Harmonic Progression
Tags: scientific calculatorcalculator techniqueCASIO calculatorarithmetic progression by calculatorgeometric progression by calculatorharmonic progression by calculator
Recent Updates
11 - Area inside a circle but outside three other externally tangent circles 02 - Time to dissipate 90% of certain radioactive substance
01 - Find how long would it take for half amount af radium to decompose Simple Chemical Conversion
01 - Thermometer reading after 6 minutes of being outside Newton's Law of Cooling
Problem 1010 and Problem 1011 | Investigation of timber reinforced by two steel channels
Problem 1009 | Width of aluminum plate reinforcement for the wood section to resist 14 kN-m moment Problem 1008 | Finding the width of steel plate reinforcement
Problem 1007 | Flexural stresses developed in the wood and steel fibers Problem 1006 | Width of fastened steel plate for balanced reinforcement
Problem 1005 | Maximum concentrated load at the midspan that the reinforced timber beam can carry Problem 1003 | Maximum stresses in wood and steel of composite beam
Problem 1004 | Increase in moment capacity due to aluminum plate reinforcement Problem 1002 | Increase in moment capacity due to steel plate reinforcement Beams with Different Materials
Chapter 10 - Reinforced Beams Volume of pyramid cut from a sphere
Amount of water inside the horizontal cylindrical tank
01 - Circle tangent to a given line and center at another given line 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.
Romel Verterra's blog Add new comment 8088 reads SPONSORED LINKS