
Concept explainers
To find: first five terms of the geometric sequence and the common ratio. Write the nth term of the sequence as a function of n .

Answer to Problem 29E
First five terms of the geometric sequence are 6,−9,272,−814,2438 respectively.
Common ratio of the sequence is −32 .
An expression for nth term of the sequence is an=6(−32)n−1 .
Explanation of Solution
Given information:
An sequence is given with first term a1=6 and a recurrence relation ak+1=−32ak .
Concept used:
The n th term of geometric sequence has the form a1=a1rn−1 where r is the common ratio of the consecutive terms of the sequence.
Every geometric sequence can be written in the following form.
a1=a1,a2=a1r,a3=a1r2,a4=a1r3,...,an=a1rn−1,..
Calculation:
Now, consider the first term.
a1=6
And
ak+1=−32ak
First five terms of the sequence can be found by substituting 1, 2, 3, 4, and 5 for kas shown:
a2=−32a1=−32×6=−9a3=−32a2=−32×(−9)=272a4=−32a3=−32×272=−814a5=−32a4=−32×(−814)=2438
Hence, the first five terms of the sequence are 6,−9,272,−814,2438 respectively.
The common ratio can be found by dividing first term by second term.
r=a2a1=−96=−32
Therefore, the common ratio is −32 .
Now, the expression for nth term can be found as
an=a1rn−1=6(−32)n−1an=6(−32)n−1
Hence, the expression for nth term of the sequence is an=6(−32)n−1 .
Chapter 8 Solutions
Precalculus with Limits: A Graphing Approach
- 1) Find the equation of the tangent line to the graph y=xe at the point (1, 1).arrow_forward3) Suppose that f is differentiable on [0, 5], and f'(x) ≤ 3 over this interval. If f(0) = −1, what is the maximum possible value of f(5)?arrow_forward2) Find the maximum value of f(x, y) = x - y on the circle x² + y² - 4x - 2y - 4 = 0.arrow_forward
- For the system consisting of the lines: and 71 = (-8,5,6) + t(4, −5,3) 72 = (0, −24,9) + u(−1, 6, −3) a) State whether the two lines are parallel or not and justify your answer. b) Find the point of intersection, if possible, and classify the system based on the number of points of intersection and how the lines are related. Show a complete solution process.arrow_forward3. [-/2 Points] DETAILS MY NOTES SESSCALCET2 7.4.013. Find the exact length of the curve. y = In(sec x), 0 ≤ x ≤ π/4arrow_forwardH.w WI M Wz A Sindax Sind dy max Утах at 0.75m from A w=6KN/M L=2 W2=9 KN/m P= 10 KN B Make the solution handwritten and not artificial intelligence because I will give a bad rating if you solve it with artificial intelligencearrow_forward
- Solve by DrWz WI P L B dy Sind Ⓡ de max ⑦Ymax dx Solve by Dr ③Yat 0.75m from A w=6KN/M L=2 W2=9 kN/m P= 10 KN Solve By Drarrow_forwardHow to find the radius of convergence for the series in the image below? I'm stuck on how to isolate the x in the interval of convergence.arrow_forwardDetermine the exact signed area between the curve g(x): x-axis on the interval [0,1]. = tan2/5 secx dx andarrow_forward
- Calculus: Early TranscendentalsCalculusISBN:9781285741550Author:James StewartPublisher:Cengage LearningThomas' Calculus (14th Edition)CalculusISBN:9780134438986Author:Joel R. Hass, Christopher E. Heil, Maurice D. WeirPublisher:PEARSONCalculus: Early Transcendentals (3rd Edition)CalculusISBN:9780134763644Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric SchulzPublisher:PEARSON
- Calculus: Early TranscendentalsCalculusISBN:9781319050740Author:Jon Rogawski, Colin Adams, Robert FranzosaPublisher:W. H. FreemanCalculus: Early Transcendental FunctionsCalculusISBN:9781337552516Author:Ron Larson, Bruce H. EdwardsPublisher:Cengage Learning





