The eighth term of a geometric progression is 32 and the fourth term is 2. Find the first term and the positive value of the common ratio
b) find the sum to infinity of a geometric progression whose first term is 5 and second term is -4
c) The nth term of an arithmetic progression is 3n +7. Find the sum of the first 40 terms
Anyone to help me solve this please
there are 4 terms from the 4th to the 8th, so r^4 = 32/2 = 16
r = ±2
if a1 = 5 and a2 = -4, then r = -4/5
S = a/(1-r) = 5/(1 + 4/5) = 25/9
if the nth term is 3n+7, then
a = 10 and d = 3
S40 = 40/2 (2*10 + 39*3) = 2740
a) To find the first term and the positive value of the common ratio of a geometric progression, we can use the formulas:
nth term = a * r^(n-1)
where nth term is the term number, a is the first term, and r is the common ratio.
We are given that the eighth term is 32:
32 = a * r^(8-1)
32 = a * r^7
We are also given that the fourth term is 2:
2 = a * r^(4-1)
2 = a * r^3
Now we have a system of two equations with two variables (a and r):
32 = a * r^7
2 = a * r^3
Divide the first equation by the second equation to eliminate a:
32/2 = (a * r^7) / (a * r^3)
16 = r^4
Taking the fourth root of both sides:
r = √16
r = 2
Substituting the value of r into the second equation to find a:
2 = a * (2^3)
2 = a * 8
a = 2/8
a = 1/4
Therefore, the first term is 1/4 and the positive value of the common ratio is 2.
b) To find the sum to infinity of a geometric progression, we can use the formula:
Sum = a / (1 - r)
where a is the first term and r is the common ratio.
Given that the first term is 5 and the second term is -4, we can find the common ratio:
-4 = 5 * r
r = -4/5
Substituting the values of a and r into the formula:
Sum = 5 / (1 - (-4/5))
Sum = 5 / (1 + 4/5)
Sum = 5 / (9/5)
Sum = 5 * (5/9)
Sum = 25/9
Therefore, the sum to infinity of the geometric progression is 25/9.
c) To find the sum of the first 40 terms of an arithmetic progression, we can use the formula:
Sum = (n/2) * (2a + (n-1)d)
where n is the number of terms, a is the first term, and d is the common difference.
Given that the nth term is 3n + 7, we can find the first term and the common difference:
a = 3(1) + 7 = 10
d = (3(n+1) + 7) - (3n + 7) = 3
Substituting the values of n, a, and d into the formula:
Sum = (40/2) * (2(10) + (40-1)(3))
Sum = 20 * (20 + 39(3))
Sum = 20 * (20 + 117)
Sum = 20 * 137
Sum = 2740
Therefore, the sum of the first 40 terms of the arithmetic progression is 2740.
a) To find the first term and the positive value of the common ratio in a geometric progression, we can use the formula for the nth term of a geometric progression:
an = a1 * r^(n-1)
where an is the nth term, a1 is the first term, r is the common ratio, and n is the position of the term.
Given that the eighth term is 32 and the fourth term is 2, we can substitute these values into the formula to form two equations:
32 = a1 * r^7 (equation 1)
2 = a1 * r^3 (equation 2)
We can divide equation 1 by equation 2 to eliminate a1:
(32 / 2) = (a1 * r^7) / (a1 * r^3)
16 = r^4
Now we can take the fourth root of both sides to solve for r:
r = ∛16
r = 2
Plugging in this value for r in equation 2, we can solve for a1:
2 = a1 * 2^3
2 = 8a1
a1 = 2 / 8
a1 = 1/4
Therefore, the first term is 1/4 and the positive value of the common ratio is 2.
b) To find the sum to infinity of a geometric progression, we need to determine whether the common ratio is between -1 and 1. If it is, the sum to infinity exists and can be calculated using the formula:
Sum = a1 / (1 - r)
In this case, the common ratio is -4 / 5, which is between -1 and 1. Therefore, the sum to infinity exists and can be calculated as follows:
Sum = 5 / (1 - (-4/5))
Sum = 5 / (1 + 4/5)
Sum = 5 / (9/5)
Sum = 5 * (5/9)
Sum = 25/9
Therefore, the sum to infinity of the geometric progression is 25/9.
c) The nth term of an arithmetic progression is given by the formula:
an = a1 + (n-1)d
where an is the nth term, a1 is the first term, d is the common difference, and n is the position of the term.
In this case, the nth term is 3n + 7. We can equate this to the formula for the nth term of an arithmetic progression:
3n + 7 = a1 + (n-1)d
Comparing the coefficients of n on both sides, we have:
3 = d
Now we know that the common difference is 3. We can substitute this value back into the formula to solve for a1:
3(1) + 7 = a1 + (1-1)(3)
10 = a1
Therefore, the first term is 10.
To find the sum of the first 40 terms, we can use the formula for the sum of an arithmetic progression:
Sum = n/2 * (first term + last term)
In this case, n = 40, the first term is 10, and the last term is given by substituting n = 40 into the formula for the nth term:
an = a1 + (n-1)d
a40 = 10 + (40-1)(3)
a40 = 10 + 39(3)
a40 = 10 + 117
a40 = 127
Substituting these values into the formula for the sum, we get:
Sum = 40/2 * (10 + 127)
Sum = 20 * 137
Sum = 2740
Therefore, the sum of the first 40 terms is 2740.