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Mathematics 21 Online
OpenStudy (anonymous):

(1) Find the sum of the infinite geometric sequence: 27, 18, 12, … (2) Find the sum of the infinite geometric sequence: 60, -40, 20, -10, …

OpenStudy (anonymous):

help please

OpenStudy (zarkon):

is there a typo for the 2nd problem

OpenStudy (anonymous):

not that i know of

OpenStudy (zarkon):

there is not a common ratio

OpenStudy (anonymous):

thats what i thought too but its not

OpenStudy (anonymous):

For (1): the common ratio is r=12/18=2/3. Since it is less than 1, the series converges. The formula is\[s=a _{1}\div(1-r)\]where \[a _{1}\]is the first term of the series, in this case 27. Thus, \[s=27\div(1-2/3)=81\]

OpenStudy (anonymous):

r u sure

OpenStudy (anonymous):

Lulu check to see if the first term is really 80.

OpenStudy (zarkon):

81 is correct

OpenStudy (anonymous):

i checked its 60

OpenStudy (zarkon):

then it is not a geometric sequence

OpenStudy (anonymous):

Zarkon is correct.

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

Definition of a geometric series/sequence is that the terms are defined by a common ratio-- any term divided by the previous should be the same fraction. In the case of (2) we have 60/-40=-3/2, but the next pair has a ratio of -40/20=-1/2 (as do the rest of the pairs given.

OpenStudy (anonymous):

ok so what do i do?

OpenStudy (anonymous):

The problem is defective. If the first term IS 80, then the sum would be\[s=80\div(1-(-1/2))=80\div1.5=160/3\]You could "fix" the problem if this is for class, and explain this solution, otherwise there is not solution.

OpenStudy (anonymous):

ok thanks so much

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