I’m practicing linear sequences and I keep tripping over the nth-term rule when the sequence doesn’t start at the first term.
Example: I’m told the 3rd term is 23 and each term goes up by 6. I wrote the rule as a_n = 23 + (n − 3)*6 because I’m “starting from the 3rd stop” and moving forward by 6 each time. The book’s answer says a_n = 6n + 5. Are these actually the same rule?
A similar thing happened with a sequence where the 1st term is −8 and the difference is +4. I wrote a_n = −8 + (n − 1)*4, but the answer key has a_n = 4n − 12. I think they match, but my brain short-circuits when I try to see it.
My question: What’s a simple, consistent way to set up the nth-term rule in these “given the k-th term and the step” situations so I don’t mix up whether it’s (n − 1), (n − 3), etc.? And can someone explain clearly why my expressions are (or aren’t) equivalent to the book’s?
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3 Responses
Great news: you’re thinking about it exactly right! For any arithmetic (step-by-step) sequence, the most bulletproof rule is a_n = a_k + d(n − k). It literally says, “start at the k-th stop, then take (n − k) steps of size d.” Expand it and you’ll see the book’s form pop out. Your first example: a_n = 23 + 6(n − 3) = 23 + 6n − 18 = 6n + 5, which matches the answer. Your second: a_n = −8 + 4(n − 1) = −8 + 4n − 4 = 4n − 12, also a match. The (n − k) isn’t a trick-it’s just counting how many steps you’re moving from the known term to the n-th term.
Another super-consistent viewpoint (my inner algebra nerd loves this) is “slope-intercept mode”: every arithmetic sequence is linear, so a_n = dn + c. Find c from any known term: c = a_k − dk. For the first example, c = 23 − 6·3 = 5, so a_n = 6n + 5. For the second, c = −8 − 4·1 = −12, so a_n = 4n − 12. Same destination, two friendly routes.
Quick worked example: say the 5th term is 2 and the difference is d = −3. Anchor method: a_n = 2 + (−3)(n − 5) = 2 − 3n + 15 = −3n + 17. Check: n = 5 gives 2; n = 4 gives 5; n = 3 gives 8-perfect steps of −3. Or slope-intercept mode: c = a_5 − 5d = 2 − 5(−3) = 17, so a_n = −3n + 17. Same rule, same pattern. Once you lock onto either recipe-anchor at k with a_n = a_k + d(n − k), or go linear with a_n = dn + (a_k − dk)-the (n − 1), (n − 3), etc. confusion disappears.
Use the consistent template a_n = a_k + (n − k)d; expanding gives a_n = dn + (a_k − kd), so for (k,d,a_k) = (3,6,23) you get 6n + 5 and for (1,4,−8) you get 4n − 12-exactly the book’s forms. Hope this helps!
Think of it like boarding at stop k: the always-right ticket is a_n = a_k + (n − k)d; expanding shows your forms match the book’s: 23 + 6(n − 3) = 6n + 5 and −8 + 4(n − 1) = 4n − 12.
Want to try one more station-hop: if a_5 = 7 and d = −3, what’s your a_n?