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What is the fifth term of the geometric sequence a1=120, a2=36, a3=10.8, a6=0.2916?
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Jan 5, 2017
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What is the fifth term of the geometric sequence a1=120, a2=36, a3=10.8, a6=0.2916?
Mathematics
middle-school
Svarog
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The formula for geometric sequence is an = a1 r^(n-1) where r is the geometric factor and n is an interger. In the sequence given, r is equal to 3/10. In this case, an = 120* (3/10)^(n-1). Solving for a5, a5 = 120* (3/10)^(5-1) = 0.972
RyanfaeScotland
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Jan 7, 2017
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first we need to find the common ratio by dividing the second term by the first term. 36/120 = 3/10
an = a1 * r^(n-1)
n = term to find = 5
a1 = first term = 120
r = common ratio = 3/10
now we sub and solve
a5 = 120 * 3/10^(5 - 1)
a5 = 120 * 3/10^4
a5 = 120 * .0081
a5 = 0.972 <=== fifth term
or we could have just done this......since we multiply by 3/10 to find the next number...
a3 = 10.8......10.8 * 3/10 = 3.24 <== 4th term
3.24 * 3/10 = 0.972 <== fifth term
Akila
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Jan 12, 2017
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