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Using Algebra Sequences Practice Questions

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Updated on Oct 3, 2011

To review these concepts, go to Using Algebra: Sequences Study Guide.

Using Algebra Sequences Practice Questions

Problems

Practice 1

Answer each of the following questions.

  1. Find the 20th term in the sequence 2, 7, 12, 17,…
  2. Find the tenth term in the sequence 10, 6, 2, –2,…
  3. Find the value of the eighth term in the sequence 5x, 6x + 1,8x, 10x – 1,…
  4. Find the value of the fifth term in the sequence 4x – 1, 2x + 4, x + 2, x – 7,…
  5. Find the value of the seventh term in the sequence 10x – 1, 9x + 1, 7x, 4x – 4,…

Practice 2

Answer each of the following questions.

  1. Find the seventh term in the sequence 4, 12, 36, 108,…
  2. Find the value of the sixth term in the sequence x – 1, 2x, 8x, 30x + 4,…
  3. Find the value of the eighth term in the sequence 15x + 8, 8x, 4x, x + 8,…
  4. Find the value of the fifth term in the sequence , x – 2, –2x, 12x,…

Solutions

Practice 1

1. The formula tn = t1 + (n – 1)d gives us the value of any term in an arithmetic sequence. The first term, t1, is 2.
The difference, d, can be found by subtracting any pair of consecutive terms: 7 – 2 = 5, so d = 5. Because we are looking for the 20th term, n = 20:
t20 = 2 + (20 – 1)5
t20 = 2 + (19)5
t20 = 2 + 95
t20 = 97
The 20th term in the sequence is 97.
2. The formula tn = t1 + (n – 1)d gives us the value of any term in an arithmetic sequence. The first term, t1, is 10.
The difference, d, can be found by subtracting any pair of consecutive terms: 6 – 10 = –4, so d = –4. Because we are looking for the tenth term, so n = 10:
t10 = 10 + (10 – 1)(–4)
t10 = 10 + (9)(–4)
t10 = 10 – 3x6
t10 = –26
The tenth term in the sequence is –26.
3. The formula tn = t1 + (n – 1)d gives us the value of any term in an arithmetic sequence. The first term, t1, is 5x.
The difference, d, can be found by subtracting any pair of consecutive terms: (6x + 1) – 5x =x + 1, so d = x + 1. Because we are looking for the eighth term, n = 8:
t8 = 5x + (8 – 1)(x + 1)
t8 = 5x + (7)(x + 1)
t8 = 5x + (7x + 7)
t8 = 12x + 7
The eighth term in the sequence is 12x + 7.
The difference between any two consecutive terms in an arithmetic sequence is always the same. The difference between the second term and the first term is x + 1. The difference between the third term and the second term is 8x – (6x + 1) = 2x – 1.
Set these two differences equal to each other and solve for x:
x + 1 = 2x – 1
1 = x – 1
2 = x
The value of x in this sequence is 2, which means that the eighth term, 12x + 7, is equal to 12(2) + 7 = 24 + 7 = 3x1.
4. The formula tn = t1 + (n – 1)d gives us the value of any term in an arithmetic sequence. The first term, t1, is 4x – 1.
The difference, d, can be found by subtracting any pair of consecutive terms: (2x + 4) – (4x – 1) = –2x + 5, so d = –2x + 5. Because we are looking for the fifth term, n = 5:
t5 = (4x – 1) + (5 – 1)(–2x + 5)
t5 = (4x – 1) + (4)(–2x + 5)
t5 = (4x – 1) – (8x + 20)
t5 = –4x + 19
The fifth term in the sequence is –4x + 19.
The difference between any two consecutive terms in an arithmetic sequence is always the same. The difference between the second term and the first term is –2x + 5. The difference between the third term and the second term is (x + 2) – (2x + 4) = –x – 2.
Set these two differences equal to each other and solve for x:
–2x + 5 = –x – 2
5 = x – 2
7 = x
The value of x in this sequence is 7, which means that the fifth term, –4x + 19, is equal to –4(7) + 19 = –28 + 19 = –9.
5. The formula tn = t1 + (n – 1)d gives us the value of any term in an arithmetic sequence. The first term, t1, is 10x – 1.
The difference, d, can be found by subtracting any pair of consecutive terms: (9x + 1) – (10x – 1) = –x + 2, so d = –x + 2. Because we are looking for the seventh term, n = 7:
t7 = (10x – 1) + (7 – 1)(–x + 2)
t7 = (10x – 1) + (6)(–x + 2)
t7 = (10x – 1) – (6x + 12)
t7 = 4x + 11
The seventh term in the sequence is 4x + 11.
The difference between any two consecutive terms in an arithmetic sequence is always the same. The difference between the second term and the first term is –x + 2. The difference between the third term and the second term is 7x – (9x + 1) = –2x – 1.
Set these two differences equal to each other and solve for x:
x + 2 = –2x – 1
x + 2 = –1
x = –3x
The value of x in this sequence is –3x, which means that the seventh term, 4x + 11, is equal to 4(–3x) + 11 = –12 + 11 = –1.
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