6. In a GP the 3rd term is equal to four times the first term and the sixth term is equal to three times the fourth term plus 32. Determine the sequence.
GP
ar^2 = 4a ----> r^2 = 4
r = ±2
ar^5 = 3ar^3 + 32
if r = +2
32a = 24a + 32
8a =32
a = 4
the sequence is 4,8,16,32,64, 128, ...
if r = -2
-32a = -24a + 32
-8a = 32
a = -4
the sequence is -4, 8, -16, 32, -64, 128, ...
Let's denote the first term of the geometric progression (GP) as "a" and the common ratio as "r".
According to the given information,
The 3rd term is equal to four times the first term:
a * r^2 = 4a (equation 1)
The 6th term is equal to three times the fourth term plus 32:
a * r^5 = 3(a * r^3) + 32 (equation 2)
To solve for the sequence, we need to find the values of "a" and "r" that satisfy both equations.
Let's solve equation 1 for "r":
a * r^2 = 4a
Divide both sides by "a":
r^2 = 4
Taking the square root of both sides:
r = ±2 (equation 3)
Now, let's substitute equation 3 into equation 2:
a * (±2)^5 = 3(a * (±2)^3) + 32
32a = 24a + 32
32a - 24a = 32
8a = 32
Divide both sides by 8:
a = 4
So, the first term "a" is 4.
Now, substitute this value of "a" into equation 1 to find the common ratio "r":
4 * r^2 = 4 * 4
r^2 = 4
Taking the square root of both sides:
r = ±2
Since "r" can be either positive 2 or negative 2, we have two possible sequences:
Sequence 1: 4, 8, 16, 32, 64, 128, ...
Sequence 2: 4, -8, 16, -32, 64, -128, ...
These are the two possible sequences that satisfy the given conditions of the geometric progression.
To determine the sequence in a geometric progression (GP), we need to find the common ratio (r) and the first term (a).
Let's solve the problem step by step:
1. Let the first term be "a" and the common ratio be "r".
So, the second term (a₂) would be ar, the third term (a₃) would be ar², the fourth term (a₄) would be ar³, and so on.
2. Given that the 3rd term (a₃) is equal to four times the first term (4a), we can write the equation:
ar² = 4a
3. Simplify the equation:
Divide both sides by "a":
r² = 4
4. Take the square root of both sides:
r = √4
r = ±2
We have two possible values for the common ratio: r = 2 or r = -2.
5. Now, we need to find the first term (a).
6. Given that the sixth term (a₆) is equal to three times the fourth term (3a₄) plus 32, we can write the equation:
ar⁵ = 3ar³ + 32
7. Simplify the equation:
Divide both sides by "a":
r⁵ = 3r³ + 32/a
8. Substitute the value of r² from step 3:
(±2)⁵ = 3(±2)³ + 32/a
Evaluate both sides:
For r = 2:
32 = 3(8) + 32/a
32 = 24 + 32/a
32a = 56
For r = -2:
-32 = 3(-8) + 32/a
-32 = -24 + 32/a
-32a = 8
9. Solve for "a":
For r = 2:
32a = 56
a = 56/32
a = 7/4
For r = -2:
-32a = 8
a = -8/-32
a = 1/4
10. The possible sequences for the given condition are:
Sequence 1: a = 7/4, r = 2
Sequence 2: a = 1/4, r = -2
Therefore, the two possible sequences for the given conditions are:
Sequence 1: 7/4, 7/2, 7, 14, 28, 56, ...
Sequence 2: 1/4, -1/2, 1, -2, 4, -8, ...