An exponential sequence (GP)of positive terms and a linear sequence (AP)have the same first term.the sum of their first terms is 3,the sum of their second terms is 3 and the sum of their third terms is 6. Find the sum of their fifth terms
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The sum of first four terms is a G.P. is 30 and that of the last four terms is 960. Is its term is 2, find the common ratio.
To find the sum of the fifth terms of both sequences, we need to determine the common ratio (r) and common difference (d) first.
Let's assume the first term of both sequences is 'a'.
In the exponential sequence (GP), the terms are given by:
a, a*r, a*r^2, a*r^3, a*r^4, ...
In the linear sequence (AP), the terms are given by:
a, a+d, a+2d, a+3d, a+4d, ...
We are given the following information:
Sum of the first terms (GP + AP) = 3
Sum of the second terms (GP + AP) = 3
Sum of the third terms (GP + AP) = 6
Using this information, we can write the following equations:
(1) a + a = 3 -> 2a = 3 -> a = 3/2
(2) a*r + (a+d) = 3 -> 3/2 * r + (3/2 + d) = 3
(3) a*r^2 + (a+d) + (a+2d) = 6 -> 3/2 * r^2 + (3/2 + d) + (3/2 + 2d) = 6
Simplifying equations (2) and (3), we get:
(2) 3/2 * r + 3/2 + d = 3 -> 3r + 3 + 2d = 6 -> 3r + 2d = 3
(3) 3/2 * r^2 + 3/2 + d + 3/2 + 2d = 6 -> 3r^2 + 4d = 6
Now, let's solve these two equations simultaneously to find the values of r and d:
From equation (2), we can solve for r:
3r = 3 - 2d
r = (3 - 2d)/3
Substituting this value of r in equation (3):
3(3 - 2d)/3^2 + 4d = 6
3(3 - 2d) + 4d = 18
9 - 6d + 4d = 18
-2d = 18 - 9
-2d = 9
d = -9/2
d = -4.5
Substituting this value of d in equation (2):
3r + 2(-4.5) = 3
3r - 9 = 3
3r = 3 + 9
3r = 12
r = 12/3
r = 4
Therefore, the common ratio (r) of the exponential sequence (GP) is 4 and the common difference (d) of the linear sequence (AP) is -4.5.
Now, we can find the terms of both sequences and calculate the sum of their fifth terms:
In the exponential sequence (GP):
First term (a) = 3/2
Fifth term = a * r^4 = (3/2) * 4^4 = 3 * 256 = 768
In the linear sequence (AP):
First term (a) = 3/2
Fifth term = a + 4d = (3/2) + 4 * (-4.5) = (3/2) - 18 = -27/2
Sum of the fifth terms = 768 + (-27/2) = 768 - 27/2 = (1536 - 27)/2 = 1509/2 = 754.5
Therefore, the sum of the fifth terms of the exponential sequence (GP) and the linear sequence (AP) is 754.5.
To find the sum of the fifth terms of the exponential sequence and linear sequence, we need to determine the common ratio (r) for the exponential sequence and the common difference (d) for the linear sequence.
Let's assume the first term of both sequences is "a".
For the exponential sequence, the terms can be written as:
Term 1 = a
Term 2 = ar (since it's geometric sequence, each term is multiplied by the common ratio)
Term 3 = ar^2
For the linear sequence, the terms can be written as:
Term 1 = a
Term 2 = a + d (since it's an arithmetic sequence, each term is increased by the common difference)
Term 3 = a + 2d
Given that the sum of their first terms is 3, we have:
a + a = 3
2a = 3
a = 3/2
Given that the sum of their second terms is 3, we have:
ar + (a + d) = 3
ar + a + d = 3
Given that the sum of their third terms is 6, we have:
ar^2 + (a + 2d) = 6
ar^2 + a + 2d = 6
Now, substitute the value of "a" from the first equation to the second and third equations:
(3/2)r + (3/2) + d = 3
(3/2)r^2 + (3/2) + 2d = 6
Multiply the equations by 2 to eliminate the fractions:
3r + 3 + 2d = 6
3r^2 + 3 + 4d = 12
Simplify the equations:
3r + 2d = 3
3r^2 + 4d = 9
We now have a system of two equations with two variables.
To solve the system, we can use substitution or elimination methods:
Let's solve using substitution method:
From the first equation, we can express d in terms of r:
d = (3 - 3r) / 2
Substituting this value in the second equation:
3r^2 + 4((3 - 3r) / 2) = 9
3r^2 + 6 - 6r = 9
3r^2 - 6r - 3 = 0
Divide the equation by 3 to simplify:
r^2 - 2r - 1 = 0
Use the quadratic formula to solve for r:
r = (-(-2) ± √((-2)^2 - 4(1)(-1))) / (2(1))
r = (2 ± √(4 + 4)) / 2
r = (2 ± √8) / 2
Simplify the square root:
r = (2 ± 2√2) / 2
Simplify further:
r = 1 ± √2
We have two possible values for r: 1 + √2 and 1 - √2
Now, substitute these values back into the first equation to find the corresponding values of d.
For r = 1 + √2:
3(1 + √2) + 2d = 3
3 + 3√2 + 2d = 3
2d = 0 - 3 - 3√2
2d = -3 - 3√2
d = (-3 - 3√2) / 2
For r = 1 - √2:
3(1 - √2) + 2d = 3
3 - 3√2 + 2d = 3
2d = 0 - 3 + 3√2
2d = -3 + 3√2
d = (-3 + 3√2) / 2
Now that we have found the values of r and d, we can find the fifth term of each sequence and calculate their sum.
For the exponential sequence:
Term 1 = a = 3/2
Term 2 = ar = (3/2)(1 + √2)
Term 3 = ar^2 = (3/2)(1 + √2)^2
To find the fifth term of the exponential sequence, we can use the general formula for a term of a geometric sequence:
Term 5 = ar^4 = (3/2)(1 + √2)^4
For the linear sequence:
Term 1 = a = 3/2
Term 2 = a + d = (3/2) + (-3 + 3√2)/2
Term 3 = a + 2d = (3/2) + (2(-3 + 3√2))/2
To find the fifth term of the linear sequence, we can use the general formula for a term of an arithmetic sequence:
Term 5 = a + 4d = (3/2) + 4((-3 + 3√2)/2)
Finally, calculate the sum of the fifth terms of both sequences:
Sum of the fifth terms = Term 5 (exponential sequence) + Term 5 (linear sequence)