CHAPTER 1 PROGRESSIONS Arithmetic Progressions
Carl Friedrich Gauss
Mathematics is the queen of the science and arithmetic is the queen of mathematics
1 + 2 + 3 +……+ 98 + 99 + 100
The answer is 5050.
1 + 2 + 3 + …..+ 98 + 99 + 100
n S n = [ a + Tn ] 2 n
a
Tn
100 [1 + 100] = 5050 2
Sum of the first n terms of an arithmetic progression
n S n = ( a + Tn ) 2 Where
or
n S n = [ 2a + ( n − 1) d ] 2
a = first term d = common difference
Tn
= nth term
Recall previous lesson ‌
Tn = a + ( n − 1) d This formula is used to determine the nth term, Tn, of an arithmetic progression. Where
a = first term d = common difference
Example 1 Find the sum of the first 20 terms of the arithmetic progression -8, -3, 2, ‌,
Solution Given: Arithmetic Progression: -8, -3, 2, … The first term, a = -8 The common difference, d = -3-(-8) = 5 The number of terms, n = 20 Using the Formula:
Sn =
n [ 2a + ( n −1) d ] 2
Therefore, sum of the first 20 terms, S 20 :
20 S 20 = [ 2(−8) + ( 20 −1)( 5) ] 2 = 10(79) = 790
Example 2 Find the sum of the arithmetic progression 2, 5, 8, 11, ‌, 59.
Solution Given: Arithmetic Progression: 2, 5, 8, 11, …, 59. The first term, a = 2 The common difference, d = 5-2 = 3 The last term, T = 59 n
a +( n −1)d =Tn
Therefore,
2 + ( n − 1) (3) = 59 2 + 3n − 3 = 59 3n = 60 n = 20
20 S 20 = [ 2 + 59] 2 S 20 = 610
n S n = [ a + Tn ] 2
REFER TO WORKSHEET 1
Question 1 Find the sum of the first 15 terms of the arithmetic progression 1, 5, 9, 13, ‌,
A:
435
C:
4359
B:
D:
534
4388
Solution Given: Arithmetic Progression: 1, 5, 9, 13, …, The first term, a = 1 The common difference, d = 5-1 = 4 The number of terms, n = 15 Using the Formula:
Sn =
n [ 2a + ( n −1) d ] 2
Therefore, sum of the first 20 terms, S15 :
15 S15 = [ 2(1) + (15 −1)( 4 ) ] 2 = 7.5(58) = 435
REFER TO WORKSHEET 1
Question 2 Find the sum of the arithmetic progression 9.8, 9.2, 8.6, 8.0, ‌, -5.2.
A:
59.8
C:
5.978
B:
D:
-519.8
-7.98
Solution Given: Arithmetic Progression: 9.8, 9.2, 8.6, 8.0, The first term, a = 9.8 The common difference, d = 9.2- (9.8) = -0.6
…, -5.2.
The last term, Tn = -5.2
a +( n −1)d =Tn
9.8 + ( n − 1) (−0.6) = −5.2 9.8 − 0.6n + 0.6 = −5.2 − 0.6n = −15.6 n = 26
Therefore, S = 26 [9.8 + ( −5.2)] 26
S 26
2 = 59.8
n S n = [ a + Tn ] 2
REFER TO WORKSHEET 2
Deriving formula for the sum of the first n terms of an arithmetic progression
1. Consider the arithmetic progression below: 3, 7, 11, 15, 19, 23 2. Complete the Table 1 below. Arranged in ascending order
3
7
11
15
19
23
Arranged in descending order
23
19
15
11
7
3
Sum
26
26
26
26
26
26
What do you notice???
REFER TO WORKSHEET 2
INSTRUCTIONS • Answer the question no. 4 • Calculate the sum of arithmetic progression ( without using the formula ). • Explain how to express the sum of the arithmetic progression a, n and l. • Thus, explain how to deduce the sum of the arithmetic progression in a, d and n.
QUIZ
Find the sum of these arithmetic progressions:
(1) 1, 7, 13,...; the first 20 terms. (2) 85, 80, 75,‌, -10.
Solution Given: Arithmetic Progression: 1, 7, 13, …, The first term, a = 1 The common difference, d = 7-1 = 6 The number of terms, n = 20 Using the Formula:
Sn =
n [ 2a + ( n −1) d ] 2
Therefore, sum of the first 20 terms, S 20 :
S 20
20 = [ 2(1) + ( 20 −1)( 6) ] 2 = 10(116) = 1160
Solution Given: Arithmetic Progression: 85, 80, 75,…, -10. The first term, a = 85 The common difference, d = 80-85 = -5 The last term, T = -10 n
a +( n −1)d =Tn
Therefore,
85 + ( n − 1) (−5) = −10 85 − 5n + 5 = −10 − 5n = −100 n = 20
20 S 20 = [85 + (−10)] 2 S 20 = 750
REFER TO SUMMARY CHART
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