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Arithmetic and Geometric Sequences (page 3 of 5) Sections: Terminology and notation, Basic examples, Arithmetic and geometric sequences, Arithmetic series, Finite and infinite geometric series The two simplest sequences to work with are arithmetic and geometric sequences. An arithmetic sequence goes from one term to the next by always adding (or subtracting) the same value. For instance, 2, 5, 8, 11, 14,... and 7, 3, –1, –5,... are arithmetic, since you add 3 and subtract 4, respectively, at each step. A geometric sequence goes from one term to the next by always multiplying (or dividing) by the same value. So 1, 2, 4, 8, 16,... and 81, 27, 9, 3, 1, 1/3,... are geometric, since you multiply by 2 and divide by 3, respectively, at each step. The number added (or subtracted) at each stage of an arithmetic sequence is called the "common difference" d, because if you subtract (find the difference of) successive terms, you'll always get this common value. The number multiplied (or divided) at each stage of a geometric sequence is called the "common ratio" r, because if you divide (find the ratio of) successive terms, you'll always get this common value. Copyright © Elizabeth Stapel 20062011 All Rights Reserved
3, 11, 19, 27, 35,... To find the common difference, I have to subtract a pair of terms. It doesn't matter which pair I pick, as long as they're right next to each other: 11 – 3 = 8 The difference is always 8, so d = 8. Then the next term is 35 + 8 = 43.
2/9, 2/3, 2, 6, 18,... To find the common ratio, I have to divide a pair of terms. It doesn't matter which pair I pick, as long as they're right next to each other: The ratio is always 3,
so r
= 3. Then the sixth term is (18)(3)
= 54 and the seventh term is (54)(3)
= 162. Since arithmetic and geometric sequences are so nice and regular, they have formulas. For arithmetic sequences, the common difference is d, and the first term a_{1} is often referred to simply as "a". Since you get the next term by adding the common difference, the value of a_{2} is just a + d. The third term is a_{3} = (a + d) + d = a + 2d. The fourth term is a_{4} = (a + 2d) + d = a + 3d. Following this pattern, the nth term a_{n} will have the form a_{n} = a + (n – 1)d. For geometric sequences, the common ratio is r, and the first term a_{1} is often referred to simply as "a". Since you get the next term by multiplying by the common ratio, the value of a_{2} is just ar. The third term is a_{3} = r(ar) = ar^{2}. The fourth term is a_{4} = r(ar^{2}) = ar^{3}. Following this pattern, the nth term a_{n} will have the form a_{n} = ar^{(}^{n}^{ – 1)}.
1/2, 1, 2, 4, 8,... The differences don't match: 2 – 1 = 1, but 4 – 2 = 2. So this isn't an arithmetic sequence. On the other hand, the ratios are the same: 2 ÷ 1 = 2, 4 ÷ 2 = 2, 8 ÷ 4 = 2. So this is a geometric sequence with common ratio r = 2 and a = 1/2. To find the tenth and nth terms, I can just plug into the formula a_{n} = ar^{(}^{n}^{ – 1)}: a_{n}
= (1/2) 2^{n}^{–1}
The nth term of an arithmetic sequence is of the form a_{n} = a + (n – 1)d. In this case, that formula gives me a_{6} = a + (6 – 1)(3/2) = 5. Solving this formula for the value of the first term of the sequence, I get a = –5/2. Then: a_{1}
= –5/2, a_{2} = –5/2 + 3/2 = –1,
a_{3} = –1 + 3/2 = 1/2,
Since a_{4} and a_{8} are four places apart, then I know from the definition of an arithmetic sequence that a_{8} = a_{4} + 4d. Using this, I can then solve for the common difference d: 65 = 93 + 4d
Also, I know that a_{4} = a + (4 – 1)d, so, using the value I just found for d, I can find the value of the first term a: 93 = a + 3(–7)
Once I have the value of the first term and the value of the common difference, I can plugnchug to find the values of the first three terms and the general form of the nth term: a_{1} = 114, a_{2} = 114 – 7 = 107, a_{3} = 107 – 7 = 100 a_{n} = 114 + (n – 1)(–7)
These two terms are 12 – 5 = 7 places apart, so, from the definition of a geometric sequence, I know that a_{12} = ( a_{5 })( r^{7 }). I can use this to solve for the value of the common ratio r: 160 = (5/4)(r^{7})
Since a_{5} = ar^{4}, then I can solve for the value of the first term a: 5/4 = a(2^{4})
= 16a Once I have the value of the first term and the value of the common ratio, I can plug each into the formulas, and find my answers: a_{n} = (5/64)2^{(}^{n}^{ – 1)} a_{26} = (5/64)(2^{25}) = 2 621 440 << Previous Top  1  2  3  4  5  Return to Index Next >>



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