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§ 11.3 Geometric Sequences and Series

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**Geometric Sequences 1,5,25,125,625… 9,-3, 1, -1/3, 1/9**

In this section, we will consider another special sequence – the geometric sequence. A sequence is a geometric sequence if each term after the first is obtained by multiplying the preceding term by a fixed nonzero number. The amount by which you multiply each time is called the common ratio r of the sequence. Consider the following geometric sequence: 1,5,25,125,625… Here, r is 5. Now consider another geometric sequence: 9,-3, 1, -1/3, 1/9 In this case, r is negative. Note that the terms of the sequence are alternating in sign. That tells us that r is negative. Here r is -1/3. Blitzer, Intermediate Algebra, 5e – Slide #2 Section 11.3

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**Definition of a Geometric Sequence**

Geometric Sequences Definition of a Geometric Sequence A geometric sequence is a sequence in which each term after the first is obtained by multiplying the preceding term by a fixed nonzero constant. The amount by which we multiply each time is called the common ratio of the sequence. Blitzer, Intermediate Algebra, 5e – Slide #3 Section 11.3

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Geometric Sequences EXAMPLE Write the first five terms of the geometric sequence with the first term and common ratio r = -10. SOLUTION The first term is -3. The second term is (-3)(-10), or 30. The third term is (30)(-10), or The fourth term is (-300)(-10), or 3000, and so on. The first five terms are 30, -300, 3000, -30,000, 300,000. Blitzer, Intermediate Algebra, 5e – Slide #4 Section 11.3

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**General Term of a Geometric Sequence**

Geometric Sequences General Term of a Geometric Sequence The nth term (the general term) of a geometric sequence with first term and common ratio r is Blitzer, Intermediate Algebra, 5e – Slide #5 Section 11.3

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Geometric Sequences EXAMPLE Find the tenth term of the geometric sequence whose first term is 2 and whose common ratio is 4. SOLUTION To find the tenth term, , we replace n in the formula with 10, with 2, and r with 4. The tenth term is 524,288. Blitzer, Intermediate Algebra, 5e – Slide #6 Section 11.3

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**Geometric Sequences in Application**

EXAMPLE The population of Texas from 1990 through 1997 is shown in the following table. Year 1990 1991 1992 1993 Population in millions 16.99 17.35 17.71 18.08 1994 1995 1996 1997 18.46 18.85 19.25 19.65 (a) Divide the population for each year by the population in the preceding year. Round to three decimal places and show that the population of Texas is increasing geometrically. (b) Write the general term of the geometric sequence describing population growth for Texas n years after 1989. Blitzer, Intermediate Algebra, 5e – Slide #7 Section 11.3

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**Geometric Sequences in Application**

CONTINUED (c) Use your model from part (b) to estimate Texas’s population, in millions, for the year According to the U.S. Census Bureau, Texas’s population in 2000 was million. How well does your geometric sequence model the actual population? SOLUTION (a) The following table shows each value of the population divided by the preceding population value. Years 1991/1990 1992/1991 1993/1992 1994/1993 Quotient of populations 1.021 1995/1994 1996/1995 1997/1996 Blitzer, Intermediate Algebra, 5e – Slide #8 Section 11.3

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**Geometric Sequences in Application**

CONTINUED Since all of the quotients are the same, and greater than 1, the sequence is increasing geometrically. (b) Since the first value of the sequence (first population value) is 16.99, Since the common ratio was determined to be (in the preceding table), r = Therefore, the general term of the geometric sequence that models the population growth of Texas since 1989 is (c) To estimate Texas’s population in the year 2000, using the model from part (b), we determine that n = 2000 – 1989 = 11. Blitzer, Intermediate Algebra, 5e – Slide #9 Section 11.3

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**Geometric Sequences in Application**

CONTINUED This is the formula for the nth term of this geometric sequence. Replace n with 11. Evaluate the exponent. Multiply. Therefore, the models estimates that in the year 2000 there were million people living in Texas. The census for that year stated that there were million people living in Texas that year. The two numbers are relatively close. The geometric sequence models the population of Texas very well. Blitzer, Intermediate Algebra, 5e – Slide #10 Section 11.3

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**The Sum of the First n Terms of a Geometric Sequence**

Geometric Sequences The Sum of the First n Terms of a Geometric Sequence The sum, , of the first n terms of a geometric sequence is given by in which is the first term and r is the common ratio Blitzer, Intermediate Algebra, 5e – Slide #11 Section 11.3

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Geometric Sequences EXAMPLE Find the sum of the first 12 terms of the geometric sequence: 3, 6, 12, 24,... SOLUTION To find the sum of the first 12 terms, , we replace n in the formula with 12. The first term, , is 3. We must find r, the common ratio. Blitzer, Intermediate Algebra, 5e – Slide #12 Section 11.3

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Geometric Sequences CONTINUED We can find the common ratio by dividing the second term of , 6, 12, 24,... by the first term. Now we are ready to find the sum of the first 12 terms of the sequence. Use the formula for the sum of the first n terms of a geometric sequence. because we want the sum of the first 12 terms. Blitzer, Intermediate Algebra, 5e – Slide #13 Section 11.3

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**Geometric Sequences Use a calculator.**

CONTINUED Use a calculator. The sum of the first 12 terms is 12,285. Equivalently, this number is the 12th partial sum of the sequence. Blitzer, Intermediate Algebra, 5e – Slide #14 Section 11.3

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**Geometric Sequences Find the following sum:**

EXAMPLE Find the following sum: SOLUTION Let’s write out a few terms in the sum. Do you see that each term after the first is obtained by multiplying the preceding term by -3? To find the sum of the 7 terms (n = 7), we need to know the first term, , and the common ratio, r. The first term is 4(-3) or -12: The common ratio is -3: r = -3. Blitzer, Intermediate Algebra, 5e – Slide #15 Section 11.3

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Geometric Sequences CONTINUED Use the formula for the sum of the first n terms of a geometric sequence. because we are adding 7 terms. Use a calculator. Thus Blitzer, Intermediate Algebra, 5e – Slide #16 Section 11.3

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**Geometric Sequences in Application**

EXAMPLE A professional basketball player signs a contract with a beginning salary of $3,000,000 for the first year and an annual increase of 4% per year beginning in the second year. That is, beginning in year 2, the athlete’s salary will be 1.04 times what it was in the previous year. What is the sum of the athlete’s salaries for the first 7 years of the contract? Round to the nearest dollar. SOLUTION The salary for the first year is $3,000,000. With a 4% raise, the second-year salary is computed as follows: Salary for year 2 = 3,000, ,000,000(0.04) = 3,000,000( ) = 3,000,000(1.04). Blitzer, Intermediate Algebra, 5e – Slide #17 Section 11.3

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**Geometric Sequences in Application**

CONTINUED Each year, the salary is 1.04 times what it was in the previous year. Thus, the salary for year 3 is 1.04 times 3,000,000(1.04), or , and so on. Therefore, since each salary is the preceding salary multiplied by 1.04, r = By the same token, the salary in the 7th year will be Since the initial salary is $3,000,000, Since we are interested in determining the sum of the salaries for the first seven years, n = 7. Blitzer, Intermediate Algebra, 5e – Slide #18 Section 11.3

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**Geometric Sequences in Application**

CONTINUED The sum of the salaries for the first 7 years is $23,694,750. Blitzer, Intermediate Algebra, 5e – Slide #19 Section 11.3

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**Value of an Annuity: Interest Compounded n Times Per Year**

Annuities Value of an Annuity: Interest Compounded n Times Per Year If P is the deposit made at the end of each compounding period for an annuity at r percent annual interest compounded n times per year, the value, A, of the annuity after t years is Blitzer, Intermediate Algebra, 5e – Slide #20 Section 11.3

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Annuities EXAMPLE To save for retirement, you decide to deposit $2500 into an IRA at the end of each year for the next 40 years. If the interest rate is 9% per year compounded annually, find the value of the IRA after 40 years. SOLUTION The annuity involves 40 year-end deposits of P = $ The interest rate is 9%: r = Because the deposits are made once a year and the interest is compounded once a year, n = 1. The number of years is 40: t = 40. We replace the variables in the formula for the value of an annuity with these numbers. Blitzer, Intermediate Algebra, 5e – Slide #21 Section 11.3

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Annuities CONTINUED The value of the IRA at the end of 40 years is approximately $844,706. Blitzer, Intermediate Algebra, 5e – Slide #22 Section 11.3

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**The Sum of an Infinite Geometric Series**

If -1 < r < 1 (equivalently, |r| < 1), then the sum of the infinite geometric series in which is the first term and r is the common ratio is given by If , the infinite series does not have a sum. Blitzer, Intermediate Algebra, 5e – Slide #23 Section 11.3

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**Geometric Series Find the sum of the infinite geometric series:**

EXAMPLE Find the sum of the infinite geometric series: SOLUTION Before finding the sum, we must find the common ratio. Because r = -1/3, the condition that |r| < 1 is met. Thus, the infinite geometric series has a sum. This is the formula for the sum of an infinite geometric series. Let , and r = -1/3. Blitzer, Intermediate Algebra, 5e – Slide #24 Section 11.3

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**Geometric Series Thus, the sum of the geometric series is 9/4.**

CONTINUED Thus, the sum of the geometric series is 9/4. Blitzer, Intermediate Algebra, 5e – Slide #25 Section 11.3

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**Geometric Series Express as a fraction in lowest terms.**

EXAMPLE Express as a fraction in lowest terms. SOLUTION Observe that is an infinite geometric series with first term 83/100 and common ratio 1/100. Because r = 1/100, the condition that |r| < 1 is met. Thus, we can use our formula to find the sum. Therefore, Blitzer, Intermediate Algebra, 5e – Slide #26 Section 11.3

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**Geometric Series An equivalent fraction for is 83/99. CONTINUED**

Blitzer, Intermediate Algebra, 5e – Slide #27 Section 11.3

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**Geometric Series in Application**

EXAMPLE How much additional spending will be generated by a $10 billion tax rebate if 60% of all income is spent? SOLUTION The total amount spent is given by the infinite geometric series 60% of 10 billion 60% of 6 billion The first term is 10 billion: = 10 billion. The common ratio is 60%, or 0.6: r = Because r = 0.6, the condition that |r| < 1 is met. Thus, we can use our formula to find the sum. Blitzer, Intermediate Algebra, 5e – Slide #28 Section 11.3

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**Geometric Series in Application**

CONTINUED Therefore, This means that the total amount spent on the $10 billion tax rebate is approximately $25 billion. To determine the additional spending that will be generated, we subtract the initial $10 billion from $25 billion. Hence, the additional spending that will be generated is $25 billion - $10 billion = $15 billion. Blitzer, Intermediate Algebra, 5e – Slide #29 Section 11.3

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**Geometric Sequences and Series**

Important to Note… We’ve looked at sequences and at series now. What is the difference? In very simple terms, the difference is that a sequence is a list and a series is a sum. In many other ways they are similar – both related to functions of the positive integers. Just don’t forget the distinction. Sequence – a list Series – a sum Blitzer, Intermediate Algebra, 5e – Slide #30 Section 11.3

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