**Given:**

The sequence is
an=12n+3
. (1)

**Definition used:**

(1) The sequence
{an}
is increasing if
an<an+1
for all
n≥1
. That is,
a1<a2<a3<...
.

(2) The sequence
{an}
is decreasing and
an>an+1
for all
n≥1
. That is,
a1>a2>a3>...
.

(3) If the sequence is either increasing or decreasing, then the sequence is called monotonic; otherwise it is not monotonic.

(4) If
{an}
is the sequence with
m≤an≤M
for
m,M∈ℕ
, then the sequence is bounded.

**Calculation:**

Check whether the sequence is increasing, decreasing or not monotonic.

Obtain the
(n+1)th
term of the sequence by substituting
n+1
for *n* in equation (1).

an+1=12(n+1)+3=12n+2+3=12n+5<12n+3

Therefore,
an>an+1
for all
n≥1
.

Since
an>an+1
for all
n≥1
and by definition (1), the sequence is decreasing.

Check whether the sequence is bounded or not bounded.

Obtain the first term of the sequence by substituting 1 for *n* in equation (1).

a1=12⋅1+3=12+3=15

Thus, the first term of the sequence is
a1=15
and
a1>a2>a3>...
.

Therefore,
an≤15
for all
n≥1
. (2)

Obtain the limit of the sequence (the value of the term
an
as *n* tends to infinity).

limn→∞an=limn→∞12n+3=1∞=0

Thus, the limit of the sequence is 0.

Therefore,
0<an
for all
n≥1
. (3)

Combine the equations (2) and (3),
0<an≤15
for all
n≥1
.

Thus, by definition (4), the sequence is bounded.

Therefore, it can be concluded that the sequence is decreasing and bounded.