Endless terms, finite total.
Find the sum
8 + 4 + 2 + 1 + …
The terms never stop, yet the total is an ordinary number.
S = a1 − r = 8 ÷ (1 − ½) = 8 ÷ ½ = 16. The sum settles because |r| < 1: each term is half the one before, so the running total closes in on 16 without ever reaching it. B (15) just adds the four terms you can see, 8 + 4 + 2 + 1 = 15, and stops; the “+ …” means the terms keep going.
Geometric means each term is the last one times something.
8
× ½ →
4
× ½ →
2
× ½ →
1
4 ÷ 8 = ½ 2 ÷ 4 = ½ 1 ÷ 2 = ½
a = 8
r = 12
Arithmetic sequences subtract to find d. Geometric ones divide to find r — different move, same idea.
Only a shrinking series can have a finite total.
8, 4, 2, 1, ½, …
terms head to zero ✓
8, 16, 32, 64, …
terms blow up — no finite sum
the test: | r | < 1
the ratio must sit strictly between −1 and 1
So the formula is allowed.
S = a1 − r
first term on top, 1 minus the ratio underneath
S = 81 − ½
Each round keeps only r of what came before, so 1 − r is the slice that gets used up. A small slice means a big total.
Bottom first.
1 − ½ = ½
one half is left
S = 8 ÷ ½
dividing by a half
S = 8 · 2
flip and multiply
S = 16
the total
Dividing by ½ is the same as doubling.
Each new term fills half of what is left.
| Terms added | Running total | Gap to 16 |
|---|---|---|
| 8 | 8 | 8 |
| + 4 | 12 | 4 |
| + 2 | 14 | 2 |
| + 1 | 15 | 1 |
| + ½ | 15.5 | 0.5 |
The gap halves every time — it never quite reaches 16.
8 + 4 + 2 + 1 = 15 — that is the fourth row of the table, not the sum. The “+ …” promises more terms, and they close the last gap of 1. Only a ÷ (1 − r) = 8 ÷ ½ = 16 counts all of them.
Split the decimal into its repeating blocks. Each block is the one before divided by 100.
0.7272… = 72100 + 7210 000 + …
a = 72100, r = 1100
S = 721001 − 1100 = 7210099100
the formula
= 7299 = 811
the 100s cancel; divide by 9
When the repeat starts late, keep the front part separate — 0.8333… with only the 3 repeating:
0.8333… = 810 + (3100 + 31000 + …)
only the 3s repeat
tail = 31001 − 110 = 390 = 130
a = 3100, r = 110
810 + 130 = 2430 + 130 = 56
add the two parts
72100 = 1825 is the decimal that stops after two places. It is only the first term of the series, not the sum.
Two infinite series: one for the falls, one for the rises. The rises start at the first rebound.
falls: 18 + 6 + 2 + … = 181 − 13 = 27
a = 18, r = 13
rises: 6 + 2 + 23 + … = 61 − 13 = 9
a = 6, the first rebound
total = 27 + 9 = 36 m
27 m counts the falls alone, and 54 m (twice the falls) counts a rise before the first drop. The correct total is twice the falls minus the first drop: 54 − 18 = 36.
Negative r. Find 45 − 15 + 5 − …
r = −15 ÷ 45 = −13
the signs alternate
S = 45 ÷ (1 − (−13)) = 45 ÷ 43
1 − r becomes 1 + 13
S = 45 × 34 = 1354
33.75, less than the first term
What is different: with r < 0 the denominator is bigger than 1, so the total is smaller than a. Using r = +13 gives 1352, which is too big.
r is the unknown. An infinite geometric series has sum 25 and first term 15. Find r.
25 = 15 ÷ (1 − r)
the same formula, read backwards
1 − r = 1525 = 35
swap 25 and 1 − r
r = 25
|25| < 1 ✓
What is different: solve for the slice 1 − r first, then for r. Always test that the r you found lies between −1 and 1.
An r with x in it. For which x does 1 + 3x + 9x² + … have a finite sum?
r = 3x ÷ 1 = 3x
and 9x² ÷ 3x = 3x ✓
|3x| < 1
the test
−13 < x < 13
divide all three parts by 3
What is different: the test becomes an inequality to solve. The ends are left out: at x = 13 the series is 1 + 1 + 1 + …, which never settles.
If the sum is given too, set the formula equal to it: 1 + 3x + 9x² + … = 4 means 1 ÷ (1 − 3x) = 4, so 1 − 3x = 14 and x = 14. Test: 3x = 34 lies between −1 and 1 ✓.
The sum and another term. An infinite geometric series has sum 16 and second term 3. Find the first term.
a = 16(1 − r)
from the sum
16(1 − r) · r = 3
second term is a · r
16r² − 16r + 3 = 0 → (4r − 1)(4r − 3) = 0
a quadratic in r
r = 14 gives a = 12; r = 34 gives a = 4
both r pass the test
What is different: two facts, two unknowns. Write a from the sum, put it into the other fact, and keep every r with |r| < 1. Check: 12 + 3 + 34 + … = 12 ÷ (34) = 16 ✓.
Work out what fraction each shape is of the one before. That fraction is r.
the four corner triangles fold exactly onto the inner square
so each square has half the area of the last
areas: 36 + 18 + 9 + …
a = 36, r = 12
S = 361 − 12 = 72
the total area
Use the ratio for the quantity asked. Each new side is the old side ÷ √2, so for sides or perimeters r = 1√2, not 12.
The infinite series turns up as a repeating decimal, a ball that bounces forever, or shapes nested inside shapes. Find the first term and the ratio of each piece to the one before, check |r| < 1, then use a ÷ (1 − r). Ten-second check: with 0 < r < 1 the total must be bigger than the first term.
a = S(1 − r) = 16 · ½ = 8
the first term comes back out
Find the sum
27 − 9 + 3 − 1 + …
The ratio is −13, so S = 27 ÷ (1 − (−13)) = 27 ÷ (43) = 814. C uses r = +13 and misses the alternating sign — a negative r makes the denominator larger, so the sum comes out smaller.
Hint: the signs alternate, so r is negative. Watch what 1 − r does when r is below zero.
