Given the terms a10=3512 and a15=316384 of a geometric sequence, find the exact value of the term a30 of the sequence.
a30=3(12)29
We first use the formula for the n th term to write a10 and a15 as follows
a10=a1×r10−1=3512
a15=a1×r15−1=316384
We now divide the terms a10 and a15 to write
a15a10=a1∗r14a1×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