72 Sequence
Type
th
n term formula
Arithmetic
(a)
12, 19, 26, ...
9, = 0
Geometric
Arithmetic
(b)
8, 16, 32, .
an
0
Geometric

Ai. Arithmetic sequence

Aii. Tn = 5 + 7n

Bi. Geometric

Bii. Tn = 8 × 2ⁿ¯¹

Step-by-step explanation:

To successfully answer the questions given above, note the following:

1. If the sequence is Arithmetic, then:

2nd – 1st = 3rd – 2nd = common difference (d)

2. If the sequence is geometric, then,

2nd / 1st = 3rd / 2nd = common ratio (r)

3. A sequence can not be arithmetic geometric at the same time.

4. The nth term of arithmetic sequence is:

Tn = a + (n – 1)d

5. The nth term of geometric sequence is:

Tn = arⁿ¯¹

A. Sequence => 12, 19, 26

i. Determination of the type of sequence.

We'll begin by calculating the common difference

1st term = 12

2nd term = 19

3rd term = 26

Common difference (d) = 2nd – 1st

d = 19 – 12 = 7

OR

d = 3rd – 2nd

d = 26 – 19 = 7

Since a common difference exist in the sequence, the sequence is arithmetic sequence.

ii. Determination of the nth term.

Common difference (d) = 7

1st term (a) = 12

nth term (Tn) =?

Tn = a + (n – 1)d

Tn = 12 + (n – 1)7

Tn = 12 + 7n – 7

Tn = 5 + 7n

B. Sequence => 8, 16, 32

Bi. Determination of the type of sequence.

Let us begin by calculating the common ratio.

1st term = 8

2nd term = 16

3rd term = 32

Common ratio (r) = 2nd / 1st

r = 16 / 8

r = 2

OR

r = 3rd / 2nd

r = 32 / 16

r = 2

Since a common ratio exist in the sequence, the sequence is geometric.

Bii. Determination of the nth term.

Common ratio(r) = 2

1st term (a) = 8

nth term =?

Tn = arⁿ¯¹

Tn = 8 × 2ⁿ¯¹

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}