The third term of a geometric progression is 16. The sum of the third term and the fourth term is 8.Find
(a)the first term and the common ratio of the progression.
(b)the sum of infinity of the progression.
To find the first term and the common ratio of the progression, we can use the formula for the nth term of a geometric progression:
\[a_n = a \times r^{(n-1)}\]
Given that the third term is 16, we can substitute the values into the formula to get:
\[16 = a \times r^{(3-1)}\]
\[16 = a \times r^{2} \quad \text{(Equation 1)}\]
We are also given that the sum of the third term and the fourth term is 8. Using the formula for the sum of the first n terms of a geometric progression, we have:
\[S_n = \frac{a \times (1 - r^n)}{1-r}\]
Substituting the values and using n = 4 (to include the third and fourth term), we get:
\[16 + 16r = \frac{a \times (1 - r^4)}{1-r}\]
\[16 (1 + r) = a \times \frac{(1 - r^4)}{(1-r)} \quad \text{(Equation 2)}\]
We have two equations (Equation 1 and Equation 2) and two unknowns (a and r). By solving these equations simultaneously, we can find the values of a and r.
(a) Solving Equation 1 and Equation 2 simultaneously:
From Equation 1, we can express a in terms of r as:
\[a = \frac{16}{r^2}\]
Substituting this value in Equation 2:
\[16 (1 + r) = \frac{16}{r^2} \times \frac{(1 - r^4)}{(1-r)}\]
Multiplying through by \(r^2 (1 - r)\):
\[16r^2 (1 + r)(1-r) = 16(1 - r^4)\]
Simplifying:
\[16r^2 (1 - r^2) = 16(1 - r^4)\]
\[16r^2 - 16r^4 = 16 - 16r^4\]
Cancelling common terms:
\[16r^2 = 16\]
Dividing by 16:
\[r^2 = 1\]
Taking the square root of both sides:
\[r = \pm 1\]
Substituting r = 1 back into Equation 1:
\[16 = a \times 1^{2}\]
\[16 = a\]
So, if r = 1, then a = 16.
Substituting r = -1 back into Equation 1:
\[16 = a \times (-1)^{2}\]
\[16 = a\]
So, if r = -1, then a = 16.
Therefore, the first term (a) can either be 16 or -16, and the common ratio (r) can be either 1 or -1.
(b) The sum of infinity of the progression:
The sum of infinity of a geometric progression exists only when the absolute value of the common ratio (|r|) is less than 1.
In this case, as we have two possible values for the common ratio (1 and -1), we need to check both options.
When r = 1, the absolute value of r is 1, which is not less than 1. Therefore, the sum of infinity does not exist in this case.
When r = -1, the absolute value of r is 1, which is not less than 1. Therefore, the sum of infinity does not exist in this case either.
Hence, the sum of infinity of the progression does not exist for both values of the common ratio (1 and -1).
To find the first term and the common ratio of the geometric progression, we need to use the information that the third term is 16 and the sum of the third term and the fourth term is 8.
(a) Let's denote the first term as 'a' and the common ratio as 'r'.
We can use the formula for the nth term of a geometric progression: Tₙ = a * r^(n-1)
Given that the third term is 16, we can substitute n = 3 and Tₙ = 16 in the formula:
16 = a * r^(3-1)
16 = a * r^2
Similarly, given that the sum of the third term and the fourth term is 8, we can substitute n = 3 and n = 4 in the formula:
16 + (a * r^(4-1)) = 8
16 + (a * r^3) = 8
a * r^3 = -8
Now we have a system of equations:
16 = a * r^2
a * r^3 = -8
To solve this system, we can divide the second equation by the first equation:
(a * r^3)/(a * r^2) = (-8)/(16)
r^(3-2) = -1/2
r = -1/2
Substituting r = -1/2 into the first equation:
16 = a * (-1/2)^2
16 = a * 1/4
64 = a
Therefore, the first term is 64 and the common ratio is -1/2.
(b) To find the sum of infinity of the geometric progression, we use the formula for the sum of an infinite geometric series: S = a / (1 - r)
Substituting the values we found, a = 64 and r = -1/2:
S = 64 / (1 - (-1/2))
S = 64 / (1 + 1/2)
S = 64 / (3/2)
S = (64 * 2) / 3
S = 128 / 3
Therefore, the sum of infinity of the geometric progression is 128/3.
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