suppose an is an arithmetic progression .with common difference 2 and Gn is geometric progression.with common ratio 3 .let Cn be a sequence given by Cn=An+Gn .if C1=8 and C3=52 then find A1and G1
Let An = x + (n-1)*2
Gn = y*3^(n-1)
Then
x+y = 8
x+2 + 3y = 52
Now just solve for x and y
Well, let me first put on my mathematician clown wig and try to solve this for you. Let's go!
We know that C1 = A1 + G1, and C3 = A3 + G3. Since A1 is the first term of the arithmetic progression with a common difference of 2, we have A3 = A1 + 2(3 - 1) = A1 + 4.
Now, let's look at C1 and C3. We're given that C1 = 8 and C3 = 52. So we can substitute these values into the equations:
C1 = A1 + G1 = 8
C3 = A3 + G3 = A1 + 4 + 3G1 = 52
Now we have a system of two equations with two unknowns (A1 and G1), but luckily we're equipped with the powers of humor!
Let's solve this and put a smile on your face, shall we?
From equation 1, we have A1 = 8 - G1. Now we can substitute this into equation 2:
52 = (8 - G1) + 4 + 3G1
Simplifying, we get:
52 = 12 + 2G1
40 = 2G1
G1 = 20
Now that we know G1, we can substitute it back into equation 1 to find A1:
8 = A1 + 20
A1 = -12
So, after some mathematical clowning around, we found that A1 = -12 and G1 = 20. Ta-da!
Please note that while I've tried to solve this problem humorously, the actual calculation is accurate.
To find A1 and G1, we can use the given information and formulas for arithmetic and geometric progressions.
We know that the arithmetic progression, An, has a common difference of 2, so we can write it as:
An = a + (n-1)d
Where 'a' is the first term, 'n' is the term number, and 'd' is the common difference.
Similarly, the geometric progression, Gn, has a common ratio of 3, so we can write it as:
Gn = g * (r^(n-1))
Where 'g' is the first term, 'n' is the term number, and 'r' is the common ratio.
We are given that C1 = 8 and C3 = 52, so we can substitute these values into the formula for Cn: Cn = An + Gn.
For C1 = 8:
8 = A1 + G1
For C3 = 52:
52 = A3 + G3
We can calculate the values of A1 and G1 by solving these two equations simultaneously.
Substituting An and Gn formulas into the equation for C1:
8 = (a + (1-1)*d) + (g * (r^(1-1)))
8 = a + g
Substituting An and Gn formulas into the equation for C3:
52 = (a + (3-1)*d) + (g * (r^(3-1)))
52 = (a + 2d) + (g * (r^2))
Now, we have a system of two equations:
8 = a + g ...(1)
52 = (a + 2d) + (g * r^2) ...(2)
We can solve these equations using substitution or elimination method.
Using elimination method, we can subtract equation (1) from equation (2) to eliminate 'g' variable:
(52 - 8) = (a + 2d) + (g * r^2) - (a + g)
44 = a + 2d - a - g + g * r^2
44 = 2d + g * r^2
Since we are given that the common difference (d) of the arithmetic progression is 2 and the common ratio (r) of the geometric progression is 3, we can substitute these values into the equation:
44 = 2 * 2 + g * 3^2
44 = 4 + 9g
9g = 40
g = 40/9
Now, substitute the value of 'g' back into equation (1) to find 'a':
8 = a + (40/9)
a = 8 - (40/9)
a = 72/9 - 40/9
a = 32/9
Therefore, A1 = 32/9 and G1 = 40/9.
To solve this problem, we need to find the values of A1 and G1, which represent the first terms of the arithmetic progression (AP) and geometric progression (GP), respectively.
Let's start by finding the value of C1 using the given information. We know that C1 = A1 + G1 and that C1 = 8. Therefore, we have the equation:
A1 + G1 = 8 ----(Equation 1)
Next, let's find the value of C3 using the given information. We know that C3 = A3 + G3 and that C3 = 52. Since we are given that the common difference of the arithmetic progression is 2, we can find A3:
A3 = A1 + (3 - 1) * 2 = A1 + 4.
Now, we can rewrite the equation for C3:
A1 + 4 + G3 = 52 ----(Equation 2)
Since Cn = An + Gn, we can rewrite G3 as G1 * (3 - 1) = 2G1. Substituting this into Equation 2, we get:
A1 + 4 + 2G1 = 52.
Now, we have a system of two equations with two variables (A1 and G1):
A1 + G1 = 8 ----(Equation 1)
A1 + 4 + 2G1 = 52.
We can solve this system of equations to find the values of A1 and G1.
Subtracting Equation 1 from Equation 2, we get:
(A1 + 4 + 2G1) - (A1 + G1) = 52 - 8
Simplifying the equation, we have:
3G1 + 4 = 44
Subtracting 4 from both sides of the equation, we get:
3G1 = 40
Dividing both sides of the equation by 3, we find:
G1 = 40/3 = 13.33 (approximately)
Now, substituting the value of G1 back into Equation 1, we can find the value of A1:
A1 + 13.33 = 8
Subtracting 13.33 from both sides, we get:
A1 = 8 - 13.33 = -5.33 (approximately)
Therefore, the values of A1 and G1 are approximately -5.33 and 13.33, respectively.