Tyler has an infinite geometric series with sum 10. He increases the first term of his sequence by 4 and swiftly changes the subsequent terms so that the common ratio remains the same, creating a new geometric series with sum 15. Compute the common ratio of Tyler's series.
Solution
Solution:
Let a and r be the first term and common ratio of the original series, respectively. Then 1−ra=10 and 1−ra+4=15. Dividing these equations, we get that
aa+4=1015⟹a=8
Solving for r with 1−ra=1−r8=10 gives r=51.
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