Sequences and Series Worksheets

Sequences and Series Worksheets
• Page 1
1.
In the geometric series - 5, - 20, - 80 , . . . , S7 = ?
 a. $\frac{5}{3}$ b. $\frac{16388}{3}$ c. - 5461 d. - 27305

Solution:

r = a2a1
[The common ratio.]

r = - 20- 5 = 4
[Substitute a2 = - 20, a1 = - 5.]

Sn = a1(1 -rn)(1 Ã¢â‚¬â€œ r)
[Sum of n terms.]

S7 = (- 5)[1 - 47][1 - 4]
[Substitute n = 7, r = 4, a1 = - 5.]

S7 = (- 5)(- 16383)- 3

S7 = - 27305

2.
What is the sum of the first 6 terms of the geometric series 1 + 4 + 16 + ... ?
 a. 1365 b. 256 c. 4098 d. 4097

Solution:

r = a2a1
[The common ratio.]

r = 4 / 1 = 4
[Substitute a2 = 4, a1 = 1.]

Sn = a1(1 -rn)(1 - r)
[Sum of n terms.]

S6 = 1(1 - 46)(1 - 4)
[Substitute n = 6, r = 4, a1 = 1.]

S6 = 1365

So, the sum of the first 6 terms of the geometric series is 1365.

3.
The common difference in the arithmetic sequence 5, 6, 7, 8, . . . is ____.
 a. - 1 b. $\frac{11}{2}$ c. 1 d. $\frac{1}{2}$

Solution:

a1 = 5
[First term.]

a2 = 6
[Second term.]

d = a2 - a1 = 6 - 5 = 1
[Find the common difference.]

So, the common difference is 1.

4.
The 19th term of the arithmetic sequence 9.0, 9.5, 10.0, 10.5, . . . is ____.
 a. 18 b. 48 c. 46 d. 29

Solution:

d = a2 - a1
[Find the common difference.]

= 9.5 - 9.0
[a2 = 9.5, a1 = 9.0.]

= 0.5

an = a1 + (n - 1)d
[Formula for the nth term.]

= 9.0 + 9 = 18
a19 = 9.0 + (19 - 1)0.5
[Substitute n = 19, a1 = 9.0, d = 0.5.]

So, the 19th term of the arithmetic sequence is 18.

5.
The first term of an arithmetic sequence is 5 and the common difference is 3. The 4th term is _____.
 a. 35 b. 14 c. 20 d. 23

Solution:

an = a1 + (n - 1)d
[Formula for the nth term.]

a10 = 5 + (4 - 1)(3)
[Substitute n = 4, a1 = 5, d = 3.]

= 5 + 9

= 14

So, the 4th term is 14.

6.
The first term of an arithmetic sequence is - 8 and the 11th term is 22. What is the 12th term ?
 a. 1 b. 13 c. 25 d. 62

Solution:

an = a1 + (n - 1) d
[Formula for the nth term.]

a11 = a1 + (11 -1) d
[Substitute: n = 11, a1 = - 8, a11 = 22.]

22 = - 8 + 10d

30 = 10d

3 = d

a12 = a1 + (12 - 1) d
[Substitute n = 12 in the formula.]

= - 8 + 11(3)

= 25

So, the 12th term is 25.

7.
If 5 times the 5th term is equal to the 6 times the 6th term of an arithmetic sequence, then its 11th term is_______
 a. 11 b. 1 c. - 1

Solution:

5a5 = 6a6
[5(5th term) = 6(6th term).]

5(a1 + 4d) = 6(a1 + 5d)
[Substitute a5 = a1 + 4d, a6 = a1 + 5d.]

5a1 + 20d = 6a1 + 30d

0 = a1 + 10d

0 = a1 + (11 - 1) d = a11

So, the 11th term is 0.

8.
The common ratio of the geometric sequence 2, 10, 50, 250, . . . is ___.
 a. 5 b. $\frac{1}{5}$ c. 10 d. $\frac{1}{10}$

Solution:

r = a2a1
[The common ratio formula.]

r = 10 / 2 = 5
[Substitute a2 = 10, a1 = 2.]

r = 5

So, the common ratio of the sequence is 5.

9.
The common ratio of the geometric sequence 3, 0.3, 0.03, 0.003... is ___.
 a. 0.1 b. 0.03 c. 3 d. 0.3

Solution:

r = a2a1
[Common ratio formula .]

r = 0.3 / 3
[Substitute a2 = 0.3, a1 = 3.]

r = 0.1

So, the common ratio of the sequence is 0.1.

10.
The sum of the first 16 terms of the arithmetic series 2 + 6 + 10 + 14 + . . . is ______.
 a. 464 b. 448 c. 512 d. 496

Solution:

d = a2 - a1 = 6 - 2 = 4
[Find d, the common difference.]

Sn = n2[2a1 + (n - 1)d]
[Use the formula.]

S16 = 16 / 2[2(2) + (16 - 1)4]
[Substitute n = 16, a1 = 2, d = 4.]

S16 = 8(4 + 60)

S16 = 512

So, the sum of the first 16 terms in the series is 512.