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					Arithmetic Sequence-a sequence in which a constant d has been added to each term to get the next term. - with general formula an = a1 + (n-1) d d – common difference d = term – previous term I. To find the d, given the terms 2, 6, 10, 14, … * use the formula for d 6-2 or 10-6 or 14-10 Ans. 4 II. To find the nth term A) Given the a1 and d *use the general formula *substitute a1 and d then simplify a1= 3 d= 5 an= 3 + (n-1)(5) an= 3 + 5n -5 Ans. an= 5n - 2 B) Given the sequence *solve for the d *use the general formula *substitute the a1 and the d, then simplify 2, 6, 10, 14, … an= 2 + (n-1)(4) an= 2 + 4n -4 Ans. an = 4n – 2 III. To identify a specific term A) Given the a1, n and d *use the general formula *substitute a1, n and d, then solve for an a1= 3, d= 5 a15=? a15= 3 + (15-1)(5) a15= 3 + (14)(5) Ans. a15= 73 B) Given the nth term *substitute n with the value and solve for an an= -2 + (n-1)(-1), a15=? a15 = -2 + (15-1)(-1) a15= -2+ (-14) Ans. a15= -16 C) Given the sequence *solve for the d *use the general formula, substitute a1 and n with the value then solve for an 5, 12, 19, …, a31=? a31 = 5 + (31-1)(7) a31= 5+ (210) Ans. a31= 215 IV. To identify the position of a term A) Given the a1, an and d *substitute in the general formula a1, an and d, then solve for n a1= 3, d= 5 an=18 18= 3 + (n-1)(5) 18= 3 + 5n -5 18=-2+5n Ans. n= 4 B) Given the nth term *substitute an with the term and solve for n an= 2 + (n-1)4, 26 26 = 2 + (n-1)(4) 26= 2+ 4n – 4 26+2=4n Ans. n= 7, 26 is in the 7th position C) Given the sequence *form the nth term of the sequence and use it in solving the n 5, 7, 9, … 27 27= 5 + (n-1)(2) 27= 5 + 2n – 2 27-3= 2n Ans. n= 12, 27 is in the 12th position V. Given non-consecutive terms A) To identify the nth term * solve for the d Term2 = Term1 + (Position2-Position1)d a3= 12 and a10= 47 47= 12 + (10-3)d 47-12=7d 47= 12 + 7d d= 5 *solve for the a1 by using the general formula *substitute an with a term and the n of that term 47= a1 + (10-1)(5) 47= a1 + 45 a1 = 2 Or 12 = a1 + (3-1)(5) 12= a1 +10 a1 = 2 *form the nth term by using the general formula, substitute a1 and d the simplify an= 2 + (n-1)(5) an= 2 + 5n -5 Ans. an = 5n – 3 B) To identify a term * solve for the d Term2 = Term1 + (Position2-Position1)d

a3= 12 and a10= 47 a15=? 47= 12 + (10-3)d 47-12=7d 47= 12 + 7d d= 5 *solve for the a1 by using the general formula *substitute an with a term and the n of that term 47= a1 + (10-1)(5) 47= a1 + 45 a1 = 2 Or 12 = a1 + (3-1)(5) 12= a1 +10 a1 = 2 *form the nth term by using the general formula, substitute a1 and d the simplify an= 2 + (n-1)(5) an= 2 + 5n -5 an = 5n – 3 *use the nth term to solve for the missing term a15= 2 + (15-1)(5) a15= 2 + (14)(5) Ans. a15= 72 C) To identify the position of a term * solve for the d Term2 = Term1 + (Position2-Position1)d a3= 12 and a10= 47 an=72 47= 12 + (10-3)d 47-12=7d 47= 12 + 7d d= 5 *solve for the a1 by using the general formula *substitute an with a term and the n of that term 47= a1 + (10-1)(5) 47= a1 + 45 a1 = 2 Or 12 = a1 + (3-1)(5) 12= a1 +10 a1 = 2 *form the nth term by using the general formula, substitute a1 and d the simplify an= 2 + (n-1)(5) an= 2 + 5n -5 an = 5n – 3 *use the nth term to solve for the position 72= 2 + (n-1)(5) 72= 2 + 5n – 5 72+3=5n Ans. n= 15 Arithmetic Means- the average of two numbers - in a sequence, these are the numbers between the first and the last terms VI. To find the means, given the sequence 5, 10, 15, 20 Ans. 10 and 15 VII. To insert required means, given two terms 3 means between 4 and 32 * the sequence is 4, __, __, __, 32 * solve for the d *4 is the a1, 32 is the an, 5 is the n since 32 will at the 5th position after the insertion of the means 32= 4 + (5-1) d 32=4 + 4d 32-4= 4d d=7 *use the value of d to find the other terms a2 = 4+7= 11 a3= 11+7=18 a4=18+7=25 Ans. 3, 11, 17, 25, 32 Arithmetic Series –the sum of the terms of a sequence (1) (2)

IX. Given the nth term an=3+(n-1)2 S14=? A) solve using formula (1) *solve for an a14=3+(14-1)2 a14=29

S14 

14 (3  29) 2

Ans. S14=224 B) solve using formula (2) *solve for a1 a1=3+(1-1)2 a1=3

S14 

Ans. S14=224

14 2(3)  (14  1)2) 2

Sn 

*no last term given

n (a1  an ) *last term given 2 n S n  2a1  (n  1)d ) 2





VIII. Given the sequence A) w/ identified last term 2, 4, 6, 8, … S3=? *use formula (1), a1=2 and an=6

S3 

B) unidentified last term 2, 4, 6, 8, … S14=? *use formula (2), a1=2,d=2 and n=14

3 (2  6) 2

Ans. S3=12

Sn 

14 2(2)  (14  1)2) 2

Ans. S14=210 IX. Given the a1 and d *use the formula (2) *substitute a1, d, and n with given values a1=3, d=4 S14=?

Sn 

14 2(3)  (14  1)4) 2

Ans. S14= 406


				
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