An exponential sequence of positive terms and a linear sequence have the same first term.the sum of their first terms is 3 the sum of their second terms is 3/2,and the sum of their third terms is 6.find their fifth terms.
Since the two sequences have the same first term, and they sum to 3, they both start with 3/2.
Let the arithmetic sequence be 3/2, 3/2 + d, 3/2 + 2d, ...
Let the geometric sequence be 3/2, 3/2 r, 3/2 r^2, ...
3/2 + d + 3/2 r = 3/2
so, d = -3/2 r
3/2 + 2d + 3/2 r^2 = 6
r = 3 or -1
d = -9/2 or 3/2
AS: 3/2, -3, -15/2, ...
GS: 3/2, 9/2, 27/2, ...
or
AS: 3/2, 3, 9/2, ...
GS: 3/2, -3/2, 3/2, ...
I assume you can make it to the 5th terms of each sequence.
Cool problem!
Sorry but I really don't get it....I'm terrible at maths that's why😥
Plssssss can I get a more specific explanation I kind of have a problem with this topic
An exponential sequence of positive terms and a linear sequence have the same first term. The sum of their first term is 3, the sum of their second term is 3/2, and the sum of their third term is 6. Find the sum of their fifth terms
Well, let's do some math while still keeping it fun! We'll call the first term of both sequences "x." Now, let's find the common ratio for the exponential sequence and the common difference for the linear sequence.
For the exponential sequence:
First term = x
Second term = xr (where "r" is the common ratio)
For the linear sequence:
First term = x
Second term = x + d (where "d" is the common difference)
Now let's tackle the given information:
The sum of their first terms is 3:
x + x = 2x = 3
The sum of their second terms is 3/2:
xr + (x + d) = xr + x + d = x(r + 1) + d = 3/2
The sum of their third terms is 6:
xr^2 + (x + 2d) = xr^2 + x + 2d = x(r^2 + 1) + 2d = 6
Now that we have a system of equations, let's solve it:
From the sum of first terms:
2x = 3
x = 3/2
Substituting x back into the second equation, we get:
3/2(r + 1) + d = 3/2
3/2r + 3/2 + d = 3/2
3/2r + d = 0
From the sum of third terms:
3/2(r^2 + 1) + 2d = 6
3/2r^2 + 3/2 + 2d = 6
3/2r^2 + 2d = 6 - 3/2
3/2r^2 + 2d = 9/2
Now, let's simplify the third equation:
3/2r^2 + d = 9/2 - 2d
Now we have two equations with two variables. We'll solve them without humor:
3/2r + d = 0
3/2r^2 + d = 9/2 - 2d
By substituting the value of "d" from the first equation:
3/2r^2 + 3/2 - 2d = 9/2 - 2d
3/2r^2 + 3/2 = 9/2
3/2r^2 = 6/2
r^2 = 2/2
r^2 = 1
r = 1 or -1
Now, if r = 1, and substituting it back into the first equation:
3/2(1) + d = 0
3/2 + d = 0
d = -3/2
If r = -1, and substituting it back into the first equation:
3/2(-1) + d = 0
-3/2 + d = 0
d = 3/2
So, the two sets of possible values are:
(r = 1, d = -3/2) or (r = -1, d = 3/2)
Now, let's find the fifth term for each set of values:
For (r = 1, d = -3/2):
The fifth term of the exponential sequence with a common ratio of 1 is: x * r^4 = (3/2) * 1^4 = 3/2
The fifth term of the linear sequence with a common difference of -3/2 is: x + 4d = (3/2) + (4 * (-3/2)) = (3/2) - 6 = -9/2
For (r = -1, d = 3/2):
The fifth term of the exponential sequence with a common ratio of -1 is: x * r^4 = (3/2) * (-1)^4 = 3/2
The fifth term of the linear sequence with a common difference of 3/2 is: x + 4d = (3/2) + (4 * (3/2)) = (3/2) + 6 = 15/2
Therefore, the possible fifth terms for the sequences are -9/2 and 15/2.
To solve this problem, we need to set up a system of equations and solve for the common first term and the common ratio (in the case of the exponential sequence) or the common difference (in the case of the linear sequence).
Let's denote the first term of both sequences as 'a' (which is the same for both) and the common ratio or common difference of the sequences as 'r' or 'd' respectively.
For the exponential sequence, the general form of the terms is given by:
a, ar, ar^2, ar^3, ...
For the linear sequence, the general form of the terms is given by:
a, a+d, a+2d, a+3d, ...
Using the given information, we can create equations:
Equation 1 (sum of first terms):
a + a = 3
2a = 3
a = 3/2
Equation 2 (sum of second terms in the exponential sequence):
ar + ar^2 = 3/2
Equation 3 (sum of second terms in the linear sequence):
a + (a + d) = 3/2
2a + d = 3/2
Equation 4 (sum of third terms in the exponential sequence):
ar^2 + ar^3 = 6
Equation 5 (sum of third terms in the linear sequence):
a + 2d + (a + 3d) = 6
2a + 5d = 6
Now we have a system of equations with unknowns 'a' and either 'r' or 'd' depending on the sequence.
To find the fifth term of the exponential sequence, we will solve for the common ratio 'r'.
Solving Equation 2 and Equation 4:
ar + ar^2 = 3/2
ar^2 + ar^3 = 6
Rearranging Equation 2:
ar^2 = 3/2 - ar (Equation 6)
Substituting Equation 6 into Equation 4:
(3/2 - ar) + ar^3 = 6
Rearranging and simplifying:
ar^3 - ar + 3/2 = 6
ar^3 - ar + 3/2 - 6 = 0
ar^3 - ar - 9/2 = 0
Similarly, we will solve for the fifth term of the linear sequence by solving for the common difference 'd'.
Solving Equation 3 and Equation 5:
2a + d = 3/2
2a + 5d = 6
Rearranging Equation 3:
d = 3/2 - 2a (Equation 7)
Substituting Equation 7 into Equation 5:
2a + 5(3/2 - 2a) = 6
Rearranging and simplifying:
2a + 15/2 - 10a = 6
-8a + 15/2 = 6
-8a = 6 - 15/2
-8a = 12/2 - 15/2
-8a = -3/2
a = (-3/2)/(-8)
a = 3/16
Finally, to find the fifth term of both sequences:
For the exponential sequence with the common ratio 'r':
a, ar, ar^2, ar^3, ar^4
For the linear sequence with the common difference 'd':
a, a+d, a+2d, a+3d, a+4d
Using the values we found earlier: a = 3/2 and d = (-3/2)/8 = -3/16.
The fifth term of the exponential sequence is:
ar^4 = (3/2)(r^4)
The fifth term of the linear sequence is:
a + 4d = (3/16) + 4(-3/16)
To calculate the actual values of the fifth terms, we need to solve for the common ratio 'r' or common difference 'd' using the system of equations we derived. Unfortunately, the equations are not linear, and solving them requires numerical methods such as approximation or iteration.