How do you write the first five terms of the geometric sequence #a_1=5, a_(k+1)=-2a_k# and determine the common ratio and write the nth term of the sequence as a function of n?

2 Answers
sjc
Sep 29, 2017

first 5 terms

#{5,-10,20,-40,80}#

common ratio

#r=-2#

nth term

#a_n=5(-2)^(n-1)#

Explanation:

#a_1=5#

#a_(k+1)=-2a_k#

#a_1=5#

#a_2=-2a_1=-2xx5=-10#

#a_3=-2a_2=-2xx-10=20#

#a_4--2a_3=-2xx20=-40#

#a_5=-2a_4=-2xx-40=80#

first 5 terms

#{5,-10,20,-40,80}#

common ratio

is #a_(k+1)/a_k=(-2a_k)/a_k#

#r=-2#

nth term of any Gp

#a_n=a_1r^(n-1)#

#a_n=5(-2)^(n-1)#

WayneDeguMan
Oct 8, 2017

#a_1 = 5# and #a_(k + 1) = -2a_k#

so, #a_2 = -2a_1#

=> #a_2 = -10#

Also, #a_3 = -2a_2#

i.e. #a_3 = 20#

so, our terms are #5, -10, 20, -40,...#

Hence, 1st term 5 and common ratio -2

Now, nth term is #ar^(n - 1)#

i.e. #5(-2)^(n - 1))#

or, #(-5/2)(-2^n)#

With n = 7 we have:

#(-5/2)(-2^7) = 320#

:)>

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