0
A geometric progression has a common ratio -1/2 and the sum of its first 3 terms is 18. Find the sum to infinity?
The sum to infinity of a geometric series is given by the formula S∞=a1/(1-r), where a1 is the first term in the series and r is found by dividing any term by the term immediately before it.
What else can I help you with?
What is the sum of the first 15 terms of an arithmetic?
For an Arithmetic Progression, Sum = 15[a + 7d].{a = first term and d = common difference} For a Geometric Progression, Sum = a[1-r^15]/(r-1).{r = common ratio }.
A geometric series has first term 4 and its sum to infinity is 5 Find the common ratio?
1/8
The sum to three terms of geometric series is 9 and its sum to infinity is 8. What could you deduce about the common ratio. Why. Find the first term and common ratio?
The geometric sequence with three terms with a sum of nine and the sum to infinity of 8 is -9,-18, and 36. The first term is -9 and the common ratio is -2.
The first and fourth terms of a geometric progression are 54 and 2 respectively Find the sum to infinity of the geometric progression?
t(1) = a = 54 t(4) = a*r^3 = 2 t(4)/t(1) = r^3 = 2/54 = 1/27 and so r = 1/3 Then sum to infinity = a/(1 - r) = 54/(1 - 1/3) = 54/(2/3) = 81.
How do you find the ratio in the geometric progression?
Divide any term, except the first, by the term before it.
What is the difference between arithmetic progression and geometric progression?
In an arithmetic progression the difference between each term (except the first) and the one before is a constant. In a geometric progression, their ratio is a constant. That is, Arithmetic progression U(n) - U(n-1) = d, where d, the common difference, is a constant and n = 2, 3, 4, ... Equivalently, U(n) = U(n-1) + d = U(1) + (n-1)*d Geometric progression U(n) / U(n-1) = r, where r, the common ratio is a non-zero constant and n = 2, 3, 4, ... Equivalently, U(n) = U(n-1)*r = U(1)*r^(n-1).
What does summation of infinite series?
The summation of a geometric series to infinity is equal to a/1-rwhere a is equal to the first term and r is equal to the common difference between the terms.
What is the formula for the geometric progression with the first 3 terms 4 2 1?
The nth term of the series is [ 4/2(n-1) ].
What is the difference of arithmetic progression to geometric progression?
In an arithmetic progression the difference between each term (except the first) and the one before is a constant. In a geometric progression, their ratio is a constant.That is,Arithmetic progressionU(n) - U(n-1) = d, where d, the common difference, is a constant and n = 2, 3, 4, ...Equivalently,U(n) = U(n-1) + d = U(1) + (n-1)*dGeometric progressionU(n) / U(n-1) = r, where r, the common ratio is a non-zero constant and n = 2, 3, 4, ...Equivalently,U(n) = U(n-1)*r = U(1)*r^(n-1).
What will be the first term and the common ratio of geometric progression whose 6th and 9th terms are 160 and 1280 respectively?
To find the first term and common ratio of a geometric progression, we can use the formula for the nth term of a geometric sequence: (a_n = a_1 \times r^{(n-1)}). Given that the 6th term is 160 and the 9th term is 1280, we can set up two equations using these values. From the 6th term, we get (a_1 \times r^5 = 160), and from the 9th term, we get (a_1 \times r^8 = 1280). By dividing the two equations, we can eliminate (a_1) and solve for the common ratio (r).
What does Geometric Series represent?
A geometric series represents the partial sums of a geometric sequence. The nth term in a geometric series with first term a and common ratio r is:T(n) = a(1 - r^n)/(1 - r)
What is the 7th term in the geometric sequence whose first term is 5 and the common ratio is -2?
Find the 7th term of the geometric sequence whose common ratio is 1/2 and whose first turn is 5
