1.
In the geometric series - 5, - 20, - 80 , . . . , S7 = ?
Solution:
r = a2a1
[The common ratio.]
r = - 20- 5 = 4
[Substitute a2 = - 20, a1 = - 5.]
Sn = a1(1 -rn)(1 – r)
[Sum of n terms.]
S7 = (- 5)[1 - 47][1 - 4]
[Substitute n = 7, r = 4, a1 = - 5.]
S7 = (- 5)(- 16383)- 3
S7 = - 27305
Correct answer : (4)
2.
What is the sum of the first 6 terms of the geometric series 1 + 4 + 16 + ... ?
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.
Correct answer : (1)
3.
The common difference in the arithmetic sequence 5, 6, 7, 8, . . . is ____.
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.
Correct answer : (3)
4.
The 19th term of the arithmetic sequence 9.0, 9.5, 10.0, 10.5, . . . is ____.
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.
Correct answer : (1)
5.
The first term of an arithmetic sequence is 5 and the common difference is 3. The 4th term is _____.
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.
Correct answer : (2)
6.
The first term of an arithmetic sequence is - 8 and the 11th term is 22. What is the 12th term ?
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.
Correct answer : (3)
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_______
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.
Correct answer : (1)
8.
The common ratio of the geometric sequence 2, 10, 50, 250, . . . is ___.
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.
Correct answer : (1)
9.
The common ratio of the geometric sequence 3, 0.3, 0.03, 0.003... is ___.
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.
Correct answer : (1)
10.
The sum of the first 16 terms of the arithmetic series 2 + 6 + 10 + 14 + . . . is ______.
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.
Correct answer : (3)