Issuu on Google+

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


ARE YOU SURE YOU WANT TO QUIT ?

YES

NO


DO YOUR HOMEWORK AND BE PREPARED FOR THE NEXT LESSON…


1.1 Janjang Aritmetik