Find the first five terms of the sequence in which a1 = 6 and an = –3an – 1 – 12, if n ≥ 2.
To find the first five terms of the sequence, we can use the given formula:
an = -3(an - 1) - 12
Given that a1 = 6, let's calculate the values of a2, a3, a4, and a5 by substituting the values:
For n = 2:
a2 = -3(a2 - 1) - 12
Let's substitute a2 = x for simplicity:
x = -3(x - 1) - 12
Simplifying the equation:
x = -3x + 3 - 12
x = -3x - 9
4x = -9
x = -9/4
Therefore, a2 = -9/4.
For n = 3:
a3 = -3(a3 - 1) - 12
Let's substitute a3 = y for simplicity:
y = -3(y - 1) - 12
Simplifying the equation:
y = -3y + 3 - 12
y = -3y - 9
4y = -9
y = -9/4
Therefore, a3 = -9/4.
For n = 4:
a4 = -3(a4 - 1) - 12
Let's substitute a4 = z for simplicity:
z = -3(z - 1) - 12
Simplifying the equation:
z = -3z + 3 - 12
z = -3z - 9
4z = -9
z = -9/4
Therefore, a4 = -9/4.
For n = 5:
a5 = -3(a5 - 1) - 12
Let's substitute a5 = w for simplicity:
w = -3(w - 1) - 12
Simplifying the equation:
w = -3w + 3 - 12
w = -3w - 9
4w = -9
w = -9/4
Therefore, a5 = -9/4.
Hence, the first five terms of the sequence are:
a1 = 6
a2 = -9/4
a3 = -9/4
a4 = -9/4
a5 = -9/4
To find the first five terms of the sequence, we will use the given formula: an = -3an-1 - 12, where a1 = 6.
Step 1: Substitute n = 2 into the formula.
a2 = -3a2-1 - 12
a2 = -3a1 - 12
a2 = -3(6) - 12
a2 = -18 - 12
a2 = -30
Step 2: Substitute n = 3 into the formula.
a3 = -3a3-1 - 12
a3 = -3a2 - 12
a3 = -3(-30) - 12
a3 = 90 - 12
a3 = 78
Step 3: Substitute n = 4 into the formula.
a4 = -3a4-1 - 12
a4 = -3a3 - 12
a4 = -3(78) - 12
a4 = -234 - 12
a4 = -246
Step 4: Substitute n = 5 into the formula.
a5 = -3a5-1 - 12
a5 = -3a4 - 12
a5 = -3(-246) - 12
a5 = 738 - 12
a5 = 726
So, the first five terms of the sequence are: 6, -30, 78, -246, 726.
It probably looked like this
a1 = 6
an = -3an-1 - 12 , n≥2
So ...
a1 = 6
a2 = -3a1 - 12 = -3(6) - 12 = -30
a3 = -3a2 - 12 = -3(-30) - 12 = 78
carry on, only 2 more