Asked by calisto
the fifth term of an exponential sequence [GP] is greater than the term by 13.5 and the fourth term is greater than the third term by 9 find the common ratio and the first term.
Answers
Answered by
Reiny
I will assume it says,
"the fifth term of an exponential sequence [GP] is greater than the fourth term by 13.5"
ar^4 - ar^3 = 13.5
ar^3(r - 1) = 13.5
"the fourth term is greater than the third term by 9"
ar^3 - ar^2 = 9
ar^2(r-1) = 9
divide the first equation by the second, the a's cancel, and the r-1 cancels
r = 13.5/9 = 135/90 = 3/2
back in ar^2(r-1) = 9
a(9/4)(1/2) = 9
a = 8
check: the sequence is 8, 12, 18, 27, 40.5,60.75, ....
fourth - third = 27-18 = 9
fifth - fourth = 40.5 - 27 = 13.5
all is good
change the numbers to whatever they should be, stick with the method
"the fifth term of an exponential sequence [GP] is greater than the fourth term by 13.5"
ar^4 - ar^3 = 13.5
ar^3(r - 1) = 13.5
"the fourth term is greater than the third term by 9"
ar^3 - ar^2 = 9
ar^2(r-1) = 9
divide the first equation by the second, the a's cancel, and the r-1 cancels
r = 13.5/9 = 135/90 = 3/2
back in ar^2(r-1) = 9
a(9/4)(1/2) = 9
a = 8
check: the sequence is 8, 12, 18, 27, 40.5,60.75, ....
fourth - third = 27-18 = 9
fifth - fourth = 40.5 - 27 = 13.5
all is good
change the numbers to whatever they should be, stick with the method
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