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Sequence and Series

Sequences and Series - Student Activity Solutions

Student Activities Chapter: Sequences and Series

Solutions

1. Determine the first 5 terms of the following nth term formula:

a. Un = Sn + 3
Substitute n = 1, 2, 3, 4, and 5:
First 5 terms: U1 = 4, U2 = 5, U3 = 6, U4 = 7, U5 = 8

b. Un = 8 - 2n
Substitute n = 1, 2, 3, 4, and 5:
First 5 terms: U1 = 6, U2 = 4, U3 = 2, U4 = 0, U5 = -2

2. Determine the general formula for the nth term of each of the following arithmetic sequences:

a. 3, 7, 11, 15, 19, ...
First term (a) = 3, Common difference (d) = 4
General formula: Un = a + (n - 1) * d = 3 + (n - 1) * 4 = 4n - 1

b. 80, 77, 74, 71, 68, ...
First term (a) = 80, Common difference (d) = -3
General formula: Un = a + (n - 1) * d = 80 + (n - 1) * -3 = 83 - 3n

3. For each of the following arithmetic sequences, determine the first term, common difference, 35th term, and 52nd term!

a. 3, 7, 11, 15, 19, ...
First term (a) = 3, Common difference (d) = 4
35th term: U35 = a + (35 - 1) * d = 3 + 34 * 4 = 139
52nd term: U52 = a + (52 - 1) * d = 3 + 51 * 4 = 207

b. 80, 77, 74, 71, 68, ...
First term (a) = 80, Common difference (d) = -3
35th term: U35 = a + (35 - 1) * d = 80 + 34 * -3 = -22
52nd term: U52 = a + (52 - 1) * d = 80 + 51 * -3 = -73

4. Given the arithmetic sequence 1, 4, 7, 10, 13, determine which term has a value of 3012.

First term (a) = 1, Common difference (d) = 3
3012 = a + (n - 1) * d
3012 = 1 + (n - 1) * 3
Solving for n: n = 1005

5. Given the arithmetic sequence 95, 90, 85, 80, 75, 0. Determine the number of terms in the sequence.

First term (a) = 95, Common difference (d) = -5
Last term = 0
0 = a + (n - 1) * d
0 = 95 + (n - 1) * -5
Solving for n: n = 20

6. Given an arithmetic sequence with the 3rd and 11th terms as 7 and 23, determine:

a. First term (a)
Let common difference be d.
U3 = a + 2d = 7, U11 = a + 10d = 23
Solving these equations:
a = 1, d = 3

b. Difference
d = 3

c. 30th term
U30 = a + 29d = 1 + 29 * 3 = 88

7. Given the arithmetic sequence 2, 7, 12, ..., 502:

First term (a) = 2, Common difference (d) = 5, Last term = 502
a. Middle term
Total terms: n = (502 - 2) / 5 + 1 = 101
Middle term: U51 = 2 + 50 * 5 = 252

b. Number of terms
n = 101

8. Given the arithmetic sequence 1, 13, 25, 37, 49. Between every two consecutive terms, 5 numbers are inserted to form a new arithmetic sequence. Determine:

a. Difference between terms
Difference (D) in original sequence = 12
New difference = 12 / (5 + 1) = 2

b. Number of terms
There are 25 terms in the new sequence.

c. 24th term
U24 = 1 + (24 - 1) * 2 = 47

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