SOLUTION: 40,-20,10,-5,...:t8

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Question 35713This question is from textbook Algebra and Trigonometry
: 40,-20,10,-5,...:t8 This question is from textbook Algebra and Trigonometry

Answer by AnlytcPhil(1810)   (Show Source): You can put this solution on YOUR website!
40,-20,10,-5,...:t8

Four terms are listed.  Ask yourself the following questions 
about them:

What do you have to do to the 1st term to get the 2nd, that is,
What do you have to do to 40 to get -20?
Two possible answers are  (1) add -60; (2) multiply by -1/2

What do you have to do to the 2nd term to get the 3rd, that is,
What do you have to do to -20 to get 10?
Two possible answers are  (1) add 30; (2) multiply by -1/2

What do you have to do to the 3rd term to get the 4th, that is,
What do you have to do to 10 to get -5?
Two possible answers are  (1) add -15; (2) multiply by -1/2

The only answer common to all these is (2) multiply by -1/2

So it is an geometric sequence with t1 = 40, and r = -1/2

tn = t1(1-rn)/(1-r) 

we are looking for t8, so n=8, t1 = 40, r = -1/2

t8 = 40[1-(-1/2)8]/[1-(-1/2)]

t8 = 40(1-1/256)/(1+1/2)

t8 = 40(255/256)/(3/2)

t8 = 40(255/256)×(2/3) 

t8 = 425/16 or 26.5625

Checking with a calculator:

t1 = 40
t2 = -20
t3 = 10
t4 = -5
t5 = 2.5
t6 = -1.25
t7 = .625
t8 = -.3125

Adding them all up, 40-20+10-5+2.5-1.25+.625-.3125 = 26.5625

Edwin
AnlytcPhil@aol.com