The first and second terms of an exponential sequence (GP) are respectively the first and third terms of a linear sequence (AP) . The fourth term of the linear sequence is 10 and the sum of it's first five terms is 60. Find the first five terms of the linear sequence and the sum of the first n terms
I don't understand this question
Can you show the full wokings
ar = a+2d -- but who cares?
a+3d = 10
5/2 (2a+4d) = 60
16,14,12,...
the GP has nothing to do with the AP.
Sn = n/2 (32+(n-1)(-2)) = 17n - n^2
Let's denote the first term of the exponential sequence as 'a', and the common ratio as 'r'.
The first term of the linear sequence will then be 'a', and the common difference will be twice the common ratio of the exponential sequence, which is '2r'.
Given that the fourth term of the linear sequence is 10, we can express the fourth term as follows:
a + 3(2r) = 10
From this equation, we can solve for 'r'.
a + 6r = 10
a = 10 - 6r
To find the sum of the first five terms of the linear sequence, we can use the sum formula for an arithmetic progression:
S5 = 5/2[2a + (5-1)d]
= 5/2[2(10-6r) + 4(2r)]
= 5/2[20 - 12r + 8r]
= 5/2(20 - 4r)
Given that S5 is equal to 60, we can set up an equation and solve for 'r':
5/2(20 - 4r) = 60
50 - 10r = 120
-10r = 70
r = -7
Now that we have the value of 'r', we can substitute it back into the equation for 'a' to find its value:
a = 10 - 6r
a = 10 - 6(-7)
a = 10 + 42
a = 52
Hence, the first term of the linear sequence is 52, and the common difference is 2r = 2(-7) = -14.
The first five terms of the linear sequence can be calculated as follows:
First term (a): 52
Second term (a + d): 52 + (-14) = 38
Third term (a + 2d): 52 + 2(-14) = 24
Fourth term (a + 3d): 52 + 3(-14) = 10
Fifth term (a + 4d): 52 + 4(-14) = -4
Therefore, the first five terms of the linear sequence are 52, 38, 24, 10, -4.
To find the sum of the first n terms of the linear sequence, we can use the formula for the sum of an arithmetic progression:
Sn = n/2[2a + (n-1)d]
I hope this helps! Let me know if you have any further questions.
To solve this problem, we'll break it down into smaller steps. Let's start by finding the first and second terms of the exponential sequence (GP).
Let's assume the first term of the exponential sequence is a, and the common ratio is r. Therefore, the first term of the linear sequence (AP) is a and the third term is a*r (since it is the first and second terms of the exponential sequence, respectively).
The fourth term of the linear sequence is given as 10. Therefore, we can set up the following equation using the formula for the nth term of an arithmetic sequence:
a + 3r = 10 ...(1)
The sum of the first five terms of the linear sequence is given as 60. We can use the formula for the sum of an arithmetic sequence to write another equation:
(5/2) * (2a + 4r) = 60 ...(2)
Now we have two equations with two variables (a and r). We can solve this system of equations to find the values of a and r.
(1) Divide equation (1) by 3:
a/3 + r = 10/3
(2) Multiply equation (2) by 2/5:
(2/5) * (5/2) * (2a + 4r) = (2/5) * 60
a + 2r = 24
Now we have a new system of equations:
a/3 + r = 10/3 ...(3)
a + 2r = 24 ...(4)
To eliminate a, multiply equation (3) by 3:
3(a/3 + r) = 3(10/3)
a + 3r = 10 ...(5)
Now we have the following system of equations:
(5) a + 3r = 10
(4) a + 2r = 24
Subtract equation (4) from equation (5) to eliminate a:
(a + 3r) - (a + 2r) = 10 - 24
r = -14
Substitute the value of r into equation (4) to solve for a:
a + 2(-14) = 24
a - 28 = 24
a = 52
Now we have found the values of a and r. The first term of the linear sequence is 52, and the common difference is -14.
To find the first five terms of the linear sequence, we can use the formula for the nth term of an arithmetic sequence:
an = a + (n-1)d
Using a = 52 and d = -14, we can find the first five terms:
a1 = 52
a2 = 52 + (-14) = 38
a3 = 38 + (-14) = 24
a4 = 24 + (-14) = 10
a5 = 10 + (-14) = -4
Therefore, the first five terms of the linear sequence are 52, 38, 24, 10, and -4.
To find the sum of the first n terms of the linear sequence, we can use the formula for the sum of an arithmetic sequence:
Sn = (n/2)(2a + (n-1)d)
Using a = 52, d = -14, and n = 5, we can find the sum of the first five terms:
S5 = (5/2)(2*52 + (5-1)*(-14))
S5 = (5/2)(104 + 4*(-14))
S5 = (5/2)(104 - 56)
S5 = (5/2)(48)
S5 = 120
Therefore, the sum of the first five terms of the linear sequence is 120.