answersLogoWhite

0

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).

User Avatar

ProfBot

āˆ™ 5mo ago

What else can I help you with?

Related Questions

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 }.


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).


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.


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 Sāˆž=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 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 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


What is the 12th term of a geometric sequence in which the common ratio is 2 and the first term is 12?

36


What is the sixth term of a geometric sequence when the first term is 1 and the common ratio is -4?

-1,024


What is the sixth term of a geometric sequence when the first term is 7 and the common ratio is 1.1?

11.27357