Pssa.20090421.ade.arithmetic Series, Arithmetic Progression.solutions

HOPE Charter School 2009 PSSA Boot Camp

Tuesday, April 21, 2009

PSSA Math Reporting Categories: √ A Numbers and Operations B Measurement C Geometry √ D Algebraic Concepts √ E Data Analysis and Probability SOLUTIONS Name (legibly printed): _______________________________________________ Seminar (circle one): Ackerman Beals Capers Harvey Hickson Instructions: • Place your solution into box in front of main office by end of 4th period today. • Winner will be announced during 6th period lunch today. Winner will be randomly selected from all entries, but the answer must be correct and justified (SHOW WORK OR EXPLAIN). • Link with questions and solutions posted at http://hopecharter.blogspot.com

The first three figures of a pattern are shown. If this pattern continues, how many dots will be in the 5th figure? A 11 B 13 C 15 D 28

Solutions: Solution 1: recognize a pattern, simple incrementing The number of dots is in a sequence: 1, 3, 6 The differences between these numbers is 1 (+2) 3 (+3) 6 So this is an arithmetic sequence (not a geometric sequence which involves multiplication) that keeps incrementing by 1. Continuing 1 (+2) 3 (+3) 6 (+4) 10 (+5) 15 (NEXT

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Solution 2: recognize a general pattern, derive a formula, graphing calculator The above solution would be extremely time consuming if we were asked how many dots would be in the 200th figure, so we must derive a general formula. Looking at the sequence, the data is not linear. The following display captures are from the Texas Instruments TI-84 Plus Silver Edition.

So the growth of this sequence of dots is quadratic, that is, a second degree equation (parabolic) with equation: 1 1 y = x2 + x 2 2 1 y = x ( x + 1) 2 where x is the sequence (or iteration) and y is the number of dots.

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So 1 y ( x = 4 ) = i4 ( 4 + 1) = 10 2 1 y ( x = 5 ) = i5 ( 5 + 1) = 15 2 1 y ( x = 6 ) = i6 ( 6 + 1) = 21 2 1 y ( x = 200 ) = i200 ( 200 + 1) = 20100 2