A senior math worksheet applying geometric series: nth terms and sums, how many terms are needed to pass a target, limiting sums and recurring decimals, then real applications: the future value of regular deposits, the repayment on a reducing-balance loan, total earnings from a rising salary and the distance covered by a bouncing ball. Finishes with an extended savings-plan problem. Full working is given for every answer.
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Curriculum: Texas grade levels (TEKS). We show specific outcome codes only where they have been verified against the official curriculum document.
Completed: Geometric Series in Action: Finance, Growth and Limiting Sums
12th Grade students (typically ages 17–18) working on sequences & series. It is pitched at level 3 · challenge; an easier version is linked above.
Yes. The PDF includes a full answer sheet with the working for every question. You can also view the answers online.
About 50 minutes for 17 questions. Short, regular sessions work best: two or three a week beats one long one.
No. Click Download Free PDF and it opens immediately. It is free for personal, classroom and homeschool use.
Try 12th Grade Arithmetic and Geometric Sequences Worksheet — Level 1.
Success TutoringWorksheets12th Grade · Mathematics · Sequences & SeriesName
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Show your work. Round dollar amounts to the nearest cent at the end of each calculation, not along the way.
A geometric sequence multiplies by the same common ratio r each time: a, ar, ar2, … The nth term is Tₙ = ar[sup n − 1].
The sum of the first n terms is Sₙ = [frac a(r[sup n] − 1)/r − 1] (convenient when r > 1) or Sₙ = [frac a(1 − r[sup n])/1 − r] (convenient when r < 1).
Worked example. 3 + 6 + 12 + … to 8 terms: a = 3, r = 2, S₈ = = 3 × 255 = 765.
When −1 < r < 1, the terms shrink towards 0 and rn → 0, so the sum approaches the limiting sum S∞ = [frac a/1 − r]. If |r| ≥ 1 there is no limiting sum.
A recurring decimal is a geometric series: 0.333… = 0.3 + 0.03 + 0.003 + … with a = 0.3 and r = 0.1.
Worked example. 0.777… = = = [frac 7/9].
If $M is deposited at the end of each period at interest rate r per period, then after n deposits the first deposit has grown to M(1 + r)n − 1 and the last has earned nothing. The balance is the geometric series M + M(1 + r) + … + M(1 + r)n − 1, so FV = [frac M((1 + r)[sup n] − 1)/r].
Worked example. $100 at the end of each month at 0.5% per month for 12 months: FV = = $1233.56. Only $1200 was deposited, so $33.56 is interest.
Preview — the PDF contains the full resource and answer sheet.
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