The first term of a geometric sequence is 31(1/4),the last term is 2 and the sum of the associated series is 50(3/4).what is the common ratio and how many terms are there in the series?
a = 31 1/4 = 125/4
ar^(n-1) = 2
r^(n-1) = 8/125
Now, 8/125 = (2/5)^3
Looks to me like
r = 2/5 and n=4
and the sequence is thus
125/4, 25/2, 5, 2
I think you'll find it adds up to 203/4
To find the common ratio, we can use the formula for the nth term of a geometric sequence:
an = a1 * r^(n-1)
Here, a1 is the first term, an is the last term, and r is the common ratio.
Given information:
First term, a1 = 31/4
Last term, an = 2
Using the formula, we have:
2 = (31/4) * r^(n-1)
Next, we can use the formula for the sum of a geometric series:
Sn = a1 * (1 - r^n) / (1 - r)
Given information:
Sum of the series, Sn = 50(3/4) = 203/4
Using the formula, we have:
203/4 = (31/4) * (1 - r^n) / (1 - r)
To solve for r and n, we can set up a system of equations using the two equations we found:
Equation 1:
2 = (31/4) * r^(n-1)
Equation 2:
203/4 = (31/4) * (1 - r^n) / (1 - r)
Now, let's solve the system of equations to find the values of r and n.
To find the common ratio and the number of terms in the series, we can use the formulas for the nth term and the sum of a geometric sequence.
The nth term of a geometric sequence is given by the formula:
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 number of terms.
The sum of a geometric series is given by the formula:
Sₙ = a₁ * (1 - rⁿ) / (1 - r),
where Sₙ is the sum of the first n terms.
Given information:
a₁ = 31(1/4), last term = 2, Sₙ = 50(3/4).
We can start by finding the common ratio (r):
Since the first term (a₁) is 31(1/4) and the last term is 2, we can use the formula for the nth term to set up an equation:
2 = 31(1/4) * r^(n-1).
Similarly, the sum of the series is given as 50(3/4), so we can use the formula for the sum of a geometric series to set up another equation:
50(3/4) = 31(1/4) * (1 - rⁿ) / (1 - r).
Now we have a system of two equations with two unknowns (r and n). We can solve these equations simultaneously to find the common ratio and the number of terms.
Solving the first equation for n-1:
2 / (31(1/4)) = r^(n-1).
Simplifying:
8 / 31 = r^(n-1).
Now plugging this value into the second equation:
50(3/4) = 31(1/4) * (1 - (8 / 31)) / (1 - r).
Simplifying further:
50(3/4) = 31(1/4) * (23 / 31) / (1 - r).
Multiplying both sides by 4 to eliminate the denominators:
150 = 31 * 23 / (1 - r).
Next, multiplying both sides by (1 - r) to eliminate the fraction:
150(1 - r) = 31 * 23.
Expanding and simplifying:
150 - 150r = 713.
Now, solving for r:
-150r = 713 - 150,
-150r = 563,
r = -563/150,
r = -169/45.
Now, substituting the value of r into the equation 8/31 = r^(n-1):
8/31 = (-169/45)^(n-1).
To solve this equation for n, we can take the logarithm of both sides to eliminate the exponent:
log(8/31) = log((-169/45)^(n-1)).
Using logarithm properties:
log(8/31) = (n-1) * log(-169/45).
Solving for n:
n-1 = log(8/31) / log(-169/45).
n = 1 + (log(8/31) / log(-169/45)).
Now we have the common ratio (r = -169/45) and the number of terms (n = 1 + (log(8/31) / log(-169/45))).
Please note that the logarithm of a negative number is undefined in the real number system, so in this case, it seems that the common ratio is not valid. There might be an error in the given information or calculation, as a geometric sequence should typically have a positive common ratio.