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Question

Given the terms a10=3512 and a15=316384 of a geometric sequence, find the exact value of the term a30 of the sequence.


A

a30=3(12)19

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B

a30=3(12)59

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C

a30=3(12)29

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D

a30=3(12)39

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Solution

The correct option is C

a30=3(12)29


We first use the formula for the n th term to write a10 and a15 as follows
a10=a1×r101=3512
a15=a1×r151=316384
We now divide the terms a10 and a15 to write
a15a10=a1r14a1×r9=3163843512
Solving for r
r5=132 which gives r=12
a10=3512=a1(12)9
Solving we get a1=3
a30=3(12)29


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