I keep tripping over the exponent in geometric sequences. I know one version is a_n = a1 * r^(n-1), but sometimes people start at a0 and then it’s a_n = a0 * r^n. My brain swaps them mid-problem and I lose a factor of r somewhere.
Simple example: start at 5 with ratio 2. For the 4th term, is it 5*2^3 or 5*2^4? Seems obvious until I switch to problems where they give two random terms. If a3 = 12 and a7 = 192, I get stuck on how to set the exponents so I don’t add or drop one power when solving for r or the first term.
Bonus confusion: if the ratio is negative (like start 8 with r = -1/2), I keep messing up which terms should be positive or negative.
Is there a dead-simple rule or mental trick so I don’t have to remember two different formulas? How do you keep the indexing straight without overthinking it?
Ready to make maths more enjoyable, accessible, and fun? Join a friendly community where you can explore puzzles, ask questions, track your progress, and learn at your own pace.
By becoming a member, you unlock:
3 Responses
I use the “index difference” rule: a_n = a_k · r^(n−k), so the exponent is just how many steps you move-thus from a1=5 to a4 is 3 steps (5·2^3), from a0=5 to a4 is 4 steps (5·2^4), and from a3=12 to a7=192 gives r^(7−3)=16 so r=2 and a1=12/2^2=3.
If r is negative, the signs alternate-roughly like (−1)^n when starting at a1-so I just check parity, though I might be off by one depending on where indexing starts.
The dead-simple rule is: the exponent is just “how many jumps” you make. If you know a_k and want a_n, you multiply by r once per step, so a_n = a_k · r^(n−k). That one formula covers everything: if you start at a1, it becomes a_n = a1 · r^(n−1); if you start at a0, it becomes a_n = a0 · r^n. I literally count the jumps on my fingers. Example: start at 5 with ratio 2. If 5 = a1, then a4 = 5 · 2^(4−1) = 5 · 2^3 = 40; if 5 = a0, then a4 = 5 · 2^(4−0) = 5 · 2^4 = 80. When they give two terms, use the ratio so the starting value cancels: a_m / a_k = r^(m−k). For instance, with a3 = 12 and a7 = 192, we get r^4 = 192/12 = 16, so r = 2 (note: r = −2 also fits because the exponent 4 is even). Then back up or forward with the same jump rule, e.g., a1 = a3 / r^(3−1) = 12 / 4 = 3. For negative ratios, think “sign flips every jump”: n−k even means same sign, odd means opposite. Quick check: start 8 with r = −1/2 (taking 8 = a1), then a2 = −4, a3 = 2, a4 = −1, and in general a_n = 8 · (−1/2)^(n−1). If you stick to “exponent = number of jumps,” the indexing headaches pretty much vanish.
My dead-simple rule: a_n = a_m · r^(n−m), i.e., the exponent is just the number of “jumps” from m to n-so if 5 is a1 then a4 = 5·2^(4−1) = 40, but if 5 is a0 then a4 = 5·2^(4−0) = 80; with a3 = 12 and a7 = 192, r^(7−3) = 16 so r = 2 and a1 = 12/2^(3−1) = 3; if r < 0, the sign is (-1)^(n−m), meaning it flips each jump (even jumps positive, odd jumps negative). I used to botch this too until I drew “stepping stones” and tapped once per jump-the taps are your r’s and the sign flips on every tap when r is negative; this quick primer helped me lock it in: https://www.khanacademy.org/math/algebra/sequences/geometric-sequences/v/geometric-sequences-intro.