module 1 sequences and series

December 30, 2017 | Author: api-339611548 | Category: Sequence, Ratio, Arithmetic, Series (Mathematics), Analysis
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2017 MMC, GLEAM (Gifted Learners in Math) Support Program

MODULE 1: SEQUENCES and SERIES

In this module, you are expected to strengthen your competencies related to Sequences and Series. Specifically, you should be able to encounter items that will help you: (1) generate patterns; (2) differentiate types of sequences; (3) determine the nth term of a sequence; (4) solve for the means in a sequence given two of its nonconsecutive terms; (5) work with infinite sequences. ACTIVITY 1: Complete the table below. Disregard the third and final column if the sequence described is Fibonacci. Examples is provided for each type of sequence. Study the process of obtaining answers. Sequence given its first few terms Example: 2π‘₯ + 3, π‘₯ + 4, βˆ’π‘₯ + 5, βˆ’2π‘₯ + 6

Type of Sequence

Common Difference/ Common Ratio

Next two terms

nth term

Arithmetic

βˆ’π’™ + 𝟏

βˆ’πŸ‘π’™ + πŸ•, βˆ’πŸ’π’™ + πŸ–

(πŸ‘π’™ + 𝟐) + (βˆ’π’™ + 𝟏)𝒏

In identifying the type of sequence, the technique is to always test for a common difference (Arithmetic) first. If that does not apply, try for common ratio (Geometric), then finally, try to get the reciprocal of terms and test for a possible common difference (Harmonic). The difference or ratio should be consistent for any two consecutive terms for it to work. In the case of the example, since (π‘₯ + 4) βˆ’ (2π‘₯ + 3) = βˆ’π‘₯ + 1 and (βˆ’π‘₯ + 5) βˆ’ (π‘₯ + 4) = βˆ’π‘₯ + 1. The common difference is determined. Use the common difference or ratio to identify the next two terms in the sequence. In generalizing the pattern through the nth term, note that we should use a specific formula depending on the type of sequence determined. For Arithmetic, 𝒂𝒏 = (π’‚πŸ βˆ’ 𝒅) + 𝒅𝒏. In the first example, we subtract the common difference from the first term,

(2π‘₯ + 3) βˆ’ (βˆ’π‘₯ + 1 ) = πŸ‘π’™ + 𝟐 Then, we add to this result the product of the common difference and the variable, n which stands for the nth term in the sequence: 𝒂𝒏 = (πŸ‘π’™ + 𝟐) + (βˆ’π’™ + 𝟏)𝒏 2, -6, 18, -54

Geometric

If the sequence is geometric, the nth term is 𝒂𝒏 =

π’‚πŸ 𝒓

-3

162, -486

𝟐 βˆ’ (βˆ’πŸ‘)𝒏 πŸ‘

𝒓𝒏

In the example, divide the first term by the common ratio then multiply this with the common ratio raised to n.

𝟐 𝒂𝒏 = βˆ’ (βˆ’πŸ‘)𝒏 πŸ‘ Note that when the terms have alternating signs such as the one above, the sequence can only be geometric with a common ratio that is negative. This is one hint that you should find handy.

𝟏 𝟏 𝟏 𝟏 , , , πŸ“ πŸ• πŸ— 𝟏𝟏

Harmonic

𝟏 𝟏 , πŸπŸ‘ πŸπŸ“

2

𝟏 πŸ‘ + πŸπ’

If the sequence is harmonic, the nth term is the reciprocal of the nth term of its corresponding arithmetic sequence. Thus, 𝒂𝒏 =

1|P ag e

𝟏 (π’‚πŸβˆ’π’…)+𝒅𝒏

𝟏

𝟏

. In the example, this is 𝒂𝒏 = (πŸ“βˆ’πŸ)+πŸπ’ = πŸ‘+πŸπ’. ●VGChua●Eastern Samar Division

2017 MMC, GLEAM (Gifted Learners in Math) Support Program

MODULE 1: SEQUENCES and SERIES

Sequence given its first few terms

1.

8, 11, 14, 17

2.

2, -10, 50, -250

3.

11, 6, 1, -4

4.

1, 1, 2, 3, 5

5.

4, 5,

6.

4𝑦, 2𝑦, 0, βˆ’2𝑦

7.

256, 128, 64

8.

1

9.

π‘₯, 5π‘₯ 3 , 25π‘₯ 5

,

1

Common Difference/ Common Ratio

Type of Sequence

Next two terms

nth term

25 125 4

,

,

16

1

,

1

3 10 17 24

10. 2, π‘₯ + 3, 2π‘₯ + 4, 3π‘₯ + 5 11. 12.

2

1

,

,

2

,

1

3π‘₯ 2π‘₯ 5π‘₯ 3π‘₯ 16

, 8, 12, 18

3

13. 8, 13, 21, 34 2 1

14. 2, 1, 3 , 2 15.

5 π‘₯2

,

5

5 5π‘₯

, ,

2π‘₯ 4

8

16. 3, βˆ’6, 12, βˆ’24 17. π‘₯ + 1,

π‘₯ 2 + 2π‘₯ + 1, 2π‘₯ 2 + 3π‘₯ + 1

ACTIVITY 2: Insert the required number of means in between the terms given for each sequence below. ARITHMETIC GEOMETRIC HARMONIC 1 1 18, _____, _____, _____, βˆ’2 3, _____, _____, _____, 31 , , , , 3 47 5, _____, _____, 12.5, _____

_____, 7, _____, _____, 56

π‘Ž βˆ’ 𝑏, ________, ________,7π‘Ž βˆ’ 4𝑏

π‘š + 𝑛, __________, _________, π‘š 7 + π‘š 7 𝑛

1 , 5

,

,

,

43 15

______, _______, π‘₯ βˆ’ 1, _______, π‘₯ + 3 2|P ag e

3 , 5

,

,

,

1 135

βˆ’4, ______, _______, _______, βˆ’48

1 , 21π‘₯

,

,

1 , , 3 π‘₯ , 𝑦

,

, βˆ’5,

, ,

,

,

1 5π‘₯

2 21 π‘₯ , 4π‘₯𝑦 + 𝑦 ,

,

5 59

●VGChua●Eastern Samar Division

2017 MMC, GLEAM (Gifted Learners in Math) Support Program

MODULE 1: SEQUENCES and SERIES

Activity 3: Solve the following problems. 1. An auditorium has 20 seats on the first row, 24 seats on the second row, 28 seats on the third row, and so on and has 30 rows of seats. How many seats are in the theatre?

2. A worker employed under a contract of six months received PhP 10, 500.00 for his first month of employment. If his contract states that his salary will increase by 10% every month, how much should the worker expect by the end of his contract?

3. An auditorium has 20 seats on the first row, 24 seats on the second row, 28 seats on the third row, and so on and has 30 rows of seats. How many seats are in the theatre?

4. An auditorium has 20 seats on the first row, 24 seats on the second row, 28 seats on the third row, and so on and has 30 rows of seats. How many seats are in the theatre?

5. An auditorium has 20 seats on the first row, 24 seats on the second row, 28 seats on the third row, and so on and has 30 rows of seats. How many seats are in the theatre?

3|P ag e

●VGChua●Eastern Samar Division

2017 MMC, GLEAM (Gifted Learners in Math) Support Program

MODULE 1: SEQUENCES and SERIES

Activity 4: Find the sum of the following infinite series. 1

1 2

1 3

2

2

2

1. 1 βˆ’ + ( ) βˆ’ ( ) + β‹―

2. 1 βˆ’

4|P ag e

1 √2

1

1

2

2√2

+ βˆ’

1

+ +β‹― 4

●VGChua●Eastern Samar Division

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