The first term of a geometric sequence is 125÷4,the last term is 2, and the sum of the associated series is 203÷4. What is the common ratio and how many terms are there in the series?
note that 2=8/4
(8/4)/(125/4) = (8/125) = (2/5)^3
Looks to me like r = 2/5 and there are 4 terms:
125/4, 25/2, 5, 2
the sum is 203/4
To find the common ratio of a geometric sequence, we can use the formula:
common ratio = (last term) / (first term)
In this case, the first term is given as 125/4 and the last term is given as 2. We can substitute these values into the formula:
common ratio = 2 / (125/4)
Simplifying the expression inside the brackets, we have:
common ratio = 2 / (125/4) = 2 * (4/125) = 8/125
Therefore, the common ratio of the geometric sequence is 8/125.
To find the number of terms in the series, we can use the formula for the sum of a geometric series:
sum = (first term) * (1 - (common ratio)^n ) / (1 - common ratio)
In this case, the sum is given as 203/4, the first term is 125/4, and the common ratio is 8/125. Substituting these values into the formula, we have:
203/4 = (125/4) * (1 - (8/125)^n ) / (1 - 8/125)
Simplifying the expression inside the brackets, we have:
203 = 125 * (1 - (8/125)^n ) / (1 - 8/125)
Cross multiplying and simplifying, we have:
203 * (1 - 8/125) = 125 * (1 - (8/125)^n )
203 * (117/125) = 125 * (1 - (8/125)^n )
Simplifying further, we have:
(117/125) = 1 - (8/125)^n
Subtracting 1 from both sides, we have:
-(8/125)^n = (117/125) - 1
Simplifying, we have:
-(8/125)^n = (117/125) - (125/125)
-(8/125)^n = (117 - 125)/125
-(8/125)^n = -8/125
Taking the reciprocal of both sides, we have:
-(125/8)^n = -125/8
Since both sides are equal, we can drop the negative signs:
(125/8)^n = 125/8
Since the bases on both sides are the same, we can equate the exponents:
n = 1
Therefore, there is only 1 term in the series.
To find the common ratio and the number of terms in the series, we can use the formulas and properties of geometric sequences.
Let's denote the first term of the geometric sequence as 'a', the common ratio as 'r', and the number of terms as 'n'.
We are given that the first term is 125÷4, so a = 125÷4.
The last term is given as 2, so we can find the nth term using the formula for the nth term of a geometric sequence:
an = a * r^(n-1)
Substituting the given values, we have:
2 = (125÷4) * r^(n-1)
The sum of the series, S, is given as 203÷4, and we can calculate the sum of a geometric series using the following formula:
S = a * (1 - r^n) / (1 - r)
Substituting the given values, we have:
203÷4 = (125÷4) * (1 - r^n) / (1 - r)
To solve these equations simultaneously to find the common ratio (r) and the number of terms (n), we can eliminate r by dividing the two equations. Let's do that:
(125÷4) * (1 - r^n) / (1 - r) = 2 / [a * r^(n-1)]
(125÷4) * (1 - r^n) / (1 - r) = 2 / [(125÷4) * r^(n-1)]
Now, cross-multiply to get rid of the fractions:
(125÷4) * (1 - r^n) * [(125÷4) * r^(n-1)] = 2 * (1 - r)
After simplifying and canceling out common terms, you should obtain an equation that relates r and n. You can solve this equation using algebraic manipulation or numerical methods to find the values of r and n.
Once you find the value of the common ratio (r), you can substitute it back into any of the equations to find the number of terms (n).
Note: The calculations involved in solving the equations might be complex. Consider using a calculator or computer algebra system to solve the given equations for the common ratio and the number of terms.