The first third and ninth terms of a linear sequence are the first three terms of an exponential sequence. The 7th term of the linear sequence is 14. Find the common difference of a linear sequence the common ratio of the exponential sequence and the sum of the fifth to ninth terms of the exponential sequence
Srry idk
So we are told that
(a+2d)/a = (a+8d)/(a+2d) and a+6d = 14 or a = 14-6d
a^2 + 4ad + 4d^2 = a^2 + 8ad
leaves with 4d^2 = 4ad
d = a
We are also told thta a+6d = 14 or a = 14-6d
14-6d = d
d = 2 , then a = 2
checking:
the AS would be 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, ...
does 2, 6, 18 form a GP ?? Yes, with a common ratio of 3
The GP would be : 2, 6, 18, 54, 162, 486, 1458, 4374, 13122, ....
It all checks out.
For the AP, a = 2, d = 2
For the GP, a = 2, r = 3
the sum of the fifth to ninth terms of the exponential sequence
= ar^4 + ar^8
= a(r^4 + r^8)
= 2(81 + 6561) = 13284
or just adding the fifth and ninth term of my listed sequence
= 162 + 13122
= 13284
YEAAAHHH
Let's break down the information given step by step:
Step 1: The first, third, and ninth terms of the linear sequence are the first three terms of an exponential sequence.
Let's denote the terms of the linear sequence as follows:
First term: a₁
Second term: a₁ + d
Third term: a₁ + 2d
Seventh term: a₁ + 6d = 14
Step 2: Find the common difference (d) of the linear sequence.
From the information given, we have:
a₁ + 6d = 14
Subtracting a₁ from both sides, we get:
6d = 14 - a₁
Step 3: Determine the common ratio (r) of the exponential sequence.
Since the first, third, and ninth terms of the linear sequence are the first three terms of the exponential sequence, we can express them as follows:
First term of exponential sequence: a₁
Second term of exponential sequence: a₁ * r
Third term of exponential sequence: a₁ * r^2
Step 4: Calculate the value of r.
From the information given, we have:
a₁ * r^2 = a₁ + 2d
Substituting the value of 2d from Step 2, we get:
a₁ * r^2 = a₁ + 2(14 - a₁)
Simplifying the equation:
a₁ * r^2 = a₁ + 28 - 2a₁
a₁ * r^2 - a₁ - 28 + 2a₁ = 0
a₁ * r^2 + a₁ - 28 = 0
Step 5: Apply the quadratic formula to solve for r.
The quadratic formula is given by:
r = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 1, b = 1, and c = -28. Substituting these values, we get:
r = (-1 ± √(1^2 - 4(1)(-28))) / (2(1))
r = (-1 ± √(1 + 112)) / 2
r = (-1 ± √113) / 2
So, the common ratio (r) of the exponential sequence is (-1 ± √113) / 2.
Step 6: Find the sum of the fifth to ninth terms of the exponential sequence.
The sum of a geometric series can be calculated using the formula:
S = a * (1 - r^n) / (1 - r)
In our case, the fifth term (a₅) is given by:
a₁ * r^4
Similarly, the ninth term (a₉) is given by:
a₁ * r^8
The sum of the geometric series from the fifth to the ninth term can be expressed as:
S = a₅ * (1 - r^5) / (1 - r)
Substituting the values, we get:
S = (a₁ * r^4) * (1 - r^5) / (1 - r)
Remember, the value of r is (-1 ± √113) / 2.
So, the sum of the fifth to ninth terms of the exponential sequence is given by:
S = (a₁ * ((-1 ± √113) / 2)^4) * (1 - ((-1 ± √113) / 2)^5) / (1 - ((-1 ± √113) / 2))
To solve this problem, we need to find the common difference of the linear sequence, the common ratio of the exponential sequence, and the sum of the fifth to ninth terms of the exponential sequence.
Let's start by finding the common difference of the linear sequence. We know that the 7th term of the linear sequence is 14. In a linear sequence, the n-th term is given by the formula:
an = a1 + (n-1)d
where an is the n-th term, a1 is the first term, n is the position of the term, and d is the common difference.
Since the 7th term is 14, we can substitute the values into the formula:
14 = a1 + (7-1)d
Simplifying, we have:
14 = a1 + 6d
Now, we need another piece of information to solve for either a1 or d. It's given that the first, third, and ninth terms of the linear sequence are the first three terms of an exponential sequence. From this, we can deduce that the common difference of the linear sequence is equal to the common ratio of the exponential sequence.
Therefore, we can write:
d = r
Now, let's find the common ratio of the exponential sequence. We know that the first, third, and ninth terms of the linear sequence are the first three terms of the exponential sequence. In an exponential sequence, the n-th term is given by the formula:
an = a1 * r^(n-1)
where an is the n-th term, a1 is the first term, n is the position of the term, and r is the common ratio.
Using the given information, we have:
a3 = a1 * r^(3-1)
a1 + 2d = a1 * r^2 ....(1)
a9 = a1 * r^(9-1)
a1 + 8d = a1 * r^8 ....(2)
To eliminate a1, we can divide Equation (2) by Equation (1):
(a1 * r^8) / (a1 * r^2) = (a1 + 8d) / (a1 + 2d)
Simplifying, we get:
r^6 = (a1 + 8d) / (a1 + 2d)
Since we know that d = r, we can substitute this into the equation:
r^6 = (a1 + 8r) / (a1 + 2r)
Now, let's find the sum of the fifth to ninth terms of the exponential sequence. In an exponential sequence, the sum of the first n terms is given by the formula:
Sn = a1 * (1 - r^n) / (1 - r)
where Sn is the sum of the first n terms, a1 is the first term, r is the common ratio.
We want to find the sum of the fifth to ninth terms, so n = 9:
S9 = a1 * (1 - r^9) / (1 - r)
We need to find a relationship between S9 and the common ratio r.
Since the 7th term of the linear sequence is 14, we can substitute this into the formula for the n-th term of an exponential sequence:
14 = a1 * r^(7-1)
14 = a1 * r^6
Dividing Equation (2) by this equation, we get:
r^2 = (a1 + 8r) / 14
Now, we have two equations:
r^6 = (a1 + 8r) / (a1 + 2r)
r^2 = (a1 + 8r) / 14
We can solve these equations simultaneously to find the values of r and d.