If the sum of the first two terms of a G.P is 3 and the sum of the second and third term is -6, find the sum of the first term and common ratio.
a + ar = 3 -----> a(1+r) = 3
ar + ar^2 = -6 ---> ar(1+r) = -6
divide the 2nd equation by the first
r = -2
now sub that into the first to get a
and you are all set
To solve this problem, let's assume that the first term of the geometric progression is "a" and the common ratio is "r".
We know that the sum of the first two terms is 3, so we can write the equation:
a + ar = 3
We also know that the sum of the second and third terms is -6, so we can write another equation:
ar + ar^2 = -6
To find the sum of the first term and common ratio, we need to solve these two equations. Let's start by rearranging the first equation:
a(1 + r) = 3 ...(equation 1)
Next, let's rearrange the second equation:
ar(1 + r) = -6 ...(equation 2)
Now, divide equation 2 by equation 1 to eliminate "a(1 + r)":
-6 / 3 = ar(1 + r) / (1 + r)
-2 = ar
Now we have a value for ar, which we can substitute back into equation 1:
a(1 + r) = 3
a(1 + (-2/r)) = 3
a(1 - 2/r) = 3
a - 2a/r = 3
Now, to find the sum of the first term and common ratio, we need to solve the equation a - 2a/r = 3 for a and r. Let's multiply through by r to remove the fraction:
ar - 2a = 3r
Rearranging the equation:
ar - 3r = 2a
Factoring out "a":
a(r - 2) = 3r
Now, we can solve for a:
a = 3r / (r - 2)
To find the common ratio, we can substitute this value of a back into equation 1:
a(1 + r) = 3
(3r / (r - 2))(1 + r) = 3
Simplifying this equation:
3r + 3r^2 / (r - 2) = 3
Multiplying through by (r - 2) to get rid of the fraction:
3r(r - 2) + 3r^2 = 3(r - 2)
Expanding and rearranging the equation:
3r^2 - 6r + 3r^2 = 3r - 6
Combining like terms:
6r^2 - 3r - 6 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. After finding the values of r, we can substitute them back into equation 1 to find the corresponding values of a.
Once we have obtained the values of a and r, we can calculate the sum of the first term and common ratio by simply adding them together.
To solve this problem, we need to work with the sum of terms in a Geometric Progression (G.P).
In a G.P, the sum of the first two terms can be calculated using the formula: S = a + ar, where 'S' is the sum, 'a' is the first term, and 'r' is the common ratio.
Given that the sum of the first two terms is 3, we have: 3 = a + ar. This is equation (1).
Similarly, the sum of the second and third terms can be calculated as: S = ar + ar^2.
Given that the sum of the second and third terms is -6, we have: -6 = ar + ar^2. This is equation (2).
We now have a system of equations (equation 1 and 2) that we can solve simultaneously to find the values of 'a' and 'r'.
To solve these equations, we can use substitution or elimination method. Let's use the substitution method:
From equation (1), we have: a = 3 - ar. Now substitute this value of 'a' in equation (2):
-6 = (3 - ar)r + ar^2. Simplify and re-arrange:
-6 = 3r - ar^2 + ar. Rearrange again:
-6 = 3r - ar^2 + ar - 6r. Combine like terms:
0 = -ar^2 + (ar + 3r - 6r). Simplify:
0 = -ar^2 - 3r^2.
Now, divide through by 'r' to eliminate the 'r' term:
0 = -ar - 3r.
Factor out 'r':
0 = r(-a - 3).
Since 'r' cannot be zero, we have two possibilities:
1) r = 0, which implies that all terms in the sequence are zero. However, this will not satisfy the given information.
2) -a - 3 = 0. Rearrange this equation:
-a = 3.
Thus, a = -3.
Now that we have found the values for 'a' and 'r', the sum of the first term (a) and common ratio (r) is:
a + r = -3 + r.
However, we do not have enough information to find the exact value of 'r' in this case. The sum of the first term and common ratio is -3 + r.