45 min readNo calculatorNeeds: simultaneous equations, indices
Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — two facts give two equations
The 5th term of an arithmetic progression is 17 and the 12th term is 45. Find the first term and the common difference, and hence the sum of the first 20 terms.
1a+4d=17The 5th term is a+(5−1)d — four gaps, not five.
2a+11d=45The 12th term is a+11d.
3Subtracting: 7d=28⇒d=4From the 5th term to the 12th is 7 steps, and the total climb is 28.
4a=17−16=1Substitute d back into the simpler equation.
5S20=220[2(1)+19(4)]=10(78)=780Use (n−1)d=19×4, not 20×4.
Example 2 — dividing one term by another to find r
The 2nd term of a geometric progression is 6 and the 5th term is 48. Find a and r, and the sum of the first 8 terms.
1ar=6 and ar4=48The nth term is arn−1, so the 5th term has index 4.
2Dividing: arar4=r3=8The a cancels, which is why division is the fastest route to r.
3r=2, and then a=3From ar=6 with r=2.
4S8=2−13(28−1)=3(256−1)Since r>1, use the form with rn−1 on top to keep everything positive.
5=3×255=765Subtract the 1 before multiplying by a.
Example 3 — working backwards from a sum to infinity
The first term of a geometric progression is 12 and its sum to infinity is 18. Find the common ratio.
1S∞=1−ra=18The formula is only valid if ∣r∣<1, which we should end up confirming.
21−r12=18Substitute the first term.
31−r=1812=32Rearrange for the bracket first — it is safer than cross-multiplying immediately.
4r=1−32=31Subtract from 1. Do not stop at 32.
5∣r∣=31<1, so the sum to infinity does exist.Check the condition. Here it holds, so the answer stands.