4.1 Arithmetic and Geometric Sequences and Series

4.1 Arithmetic and Geometric Sequences and Series

A sequence is an ordered list of numbers called terms, written $a_1, a_2, a_3, \dots, a_n$, where the subscript records the term's position. A series is what you get when you add those terms together. Entrance-test items on this topic almost never ask you to prove anything; they hand you a realistic pattern — bleacher rows, a salary ladder, a depreciating jeepney, splitting bacteria — and ask for one particular term, for a running total, or for the number of terms. Every one of those items collapses to a single decision: is the pattern built by adding the same amount each step, or by multiplying by the same factor each step?

1. Arithmetic sequences

An arithmetic sequence adds a fixed number, the common difference $d$, to move from any term to the next:

$d = a_n - a_{n-1}$

In $7, 11, 15, 19, \dots$ every gap is $4$, so $d = 4$. A negative $d$ makes the list fall: $50, 44, 38, \dots$ has $d = -6$.

The $n$-th term.

$a_n = a_1 + (n-1)d$

The multiplier is $(n-1)$, not $n$, because the first term needs no step at all. For $7, 11, 15, \dots$ the twentieth term is $a_{20} = 7 + 19(4) = 83$.

Locating a known value. Set $a_n$ equal to the value and solve for $n$. Is $143$ in the list above? $143 = 7 + (n-1)4 \Rightarrow 136 = 4(n-1) \Rightarrow n = 35$, a positive whole number, so yes — it is the 35th term. A fractional $n$ would mean the value is not a term at all.

Sum of the first $n$ terms.

$S_n = \frac{n}{2}\left(a_1 + a_n\right) = \frac{n}{2}\left[2a_1 + (n-1)d\right]$

Use the first form when the last term is already known, the second when you know only $a_1$, $d$ and $n$.

Arithmetic means are the terms inserted between two given numbers so that the whole list stays arithmetic. Inserting $k$ means produces $k+2$ terms and therefore $k+1$ gaps, so $d = \dfrac{\text{last} - \text{first}}{k+1}$. Insert three arithmetic means between $7$ and $27$: $d = \frac{27-7}{4} = 5$, giving $7, 12, 17, 22, 27$. A single arithmetic mean is just the ordinary average $\frac{a+b}{2}$.

2. Geometric sequences

A geometric sequence multiplies by a fixed non-zero number, the common ratio $r$:

$r = \frac{a_n}{a_{n-1}}$

In $6, 18, 54, 162, \dots$ each term is triple the one before it, so $r = 3$ and $a_7 = 6 \cdot 3^{6} = 6(729) = 4{,}374$.

Sum of the first $n$ terms.

$S_n = \frac{a_1\left(1 - r^{n}\right)}{1 - r}, \qquad r \neq 1$

For $6, 18, 54, \dots$ the first six terms total $S_6 = \frac{6(1 - 729)}{1 - 3} = \frac{6(-728)}{-2} = 2{,}184$, which you can check by hand: $6+18+54+162+486+1458 = 2{,}184$.

Geometric means follow the same logic as arithmetic means but with multiplication: inserting $k$ geometric means creates $k+1$ jumps, so $r^{,k+1} = \dfrac{\text{last}}{\text{first}}$. Insert two geometric means between $5$ and $135$: $r^3 = 27$, so $r = 3$ and the list is $5, 15, 45, 135$. The single geometric mean of $a$ and $b$ is $\pm\sqrt{ab}$; for $4$ and $25$ it is $\pm 10$.

3. Infinite geometric series

Add a geometric sequence forever and exactly one of two things happens. If $\lvert r \rvert \ge 1$ the terms never shrink and the total runs away — the series diverges and has no finite sum. If $\lvert r \rvert < 1$ each term is a fraction of the one before, $r^n$ collapses toward zero, and the running total settles on a ceiling:

$S_\infty = \frac{a_1}{1 - r}, \qquad \lvert r \rvert < 1$

So $24 + 8 + \frac{8}{3} + \dots$ has $r = \frac13$ and $S_\infty = \frac{24}{1 - 1/3} = \frac{24}{2/3} = 36$, while $2 + 6 + 18 + \dots$ has $r = 3$ and simply has no sum. Examiners like to offer a tidy number for a divergent series, so test $\lvert r \rvert$ before computing anything.

Repeating decimals as fractions

A repeating decimal is an infinite geometric series. $0.\overline{36} = 0.36 + 0.0036 + \dots$ has $a_1 = 0.36$ and $r = 0.01$, so $S_\infty = \frac{0.36}{0.99} = \frac{36}{99} = \frac{4}{11}$. That generalises into a shortcut: a block of $k$ repeating digits sitting immediately after the decimal point equals that block over $k$ nines — $0.\overline{5} = \frac59$ and $0.\overline{123} = \frac{123}{999} = \frac{41}{333}$. When a non-repeating digit comes first, peel it off and treat the rest as a series: $0.1\overline{6} = 0.1 + \left(0.06 + 0.006 + \dots\right) = \frac{1}{10} + \frac{0.06}{0.9} = \frac{1}{10} + \frac{1}{15} = \frac{1}{6}$.

4. Harmonic and recursive sequences

A harmonic sequence is one whose reciprocals form an arithmetic sequence. In $\frac12, \frac15, \frac18, \frac1{11}, \dots$ the denominators $2, 5, 8, 11$ climb by $3$, so the next reciprocal is $14$ and the next term is $\frac1{14}$. Never hunt for a common difference among the harmonic terms themselves — flip every term first. The harmonic mean of $a$ and $b$ is $\frac{2ab}{a+b}$; for $4$ and $12$ that is $\frac{96}{16} = 6$.

A recursive (Fibonacci-style) sequence defines each term from the ones before it, most often $a_n = a_{n-1} + a_{n-2}$. Seeded with $3$ and $4$ it runs $3, 4, 7, 11, 18, 29, 47$. There is no common difference and no common ratio, so whenever both tests fail, check whether each term is the sum (or difference, or product) of its predecessors and simply extend the list.

5. Choosing the right model

Word problems announce their model in one phrase:

6. Worked example 1 — tiered seating

Problem. The bleachers of a barangay covered court in Sampaloc have 18 rows. The front row seats 14 people, and each row behind it seats 3 more than the row in front. (a) How many seats are in the last row? (b) How many seats are there altogether? (c) Which is the first row that seats more than 50 people?

1. Each row adds a fixed 3 seats, so the rows are arithmetic with $a_1 = 14$, $d = 3$, $n = 18$.
2. (a) $a_{18} = 14 + (18-1)(3) = 14 + 51 = 65$ seats in the last row.
3. (b) $S_{18} = \frac{18}{2}(14 + 65) = 9(79) = 711$ seats in total.
4. (c) Solve $14 + 3(n-1) > 50 \Rightarrow 3(n-1) > 36 \Rightarrow n > 13$. The first whole-number row is $n = 14$, seating $14 + 39 = 53$. Row 13 seats exactly 50, which is not "more than 50" — that strict-inequality boundary is the trap.

7. Worked example 2 — depreciation and a pledge ceiling

Problem. (a) A second-hand jeepney bought for ₱480,000 loses 25% of its current value every year. What is it worth at the start of year 5, and how much value has it lost in total? (b) A parish fundraising drive collects ₱64,000 on its first day, and on each following day collects three-quarters of the previous day's amount. How much comes in over the first three days, and what total can the drive never exceed?

(a) Losing 25% means keeping 75%, so $r = 0.75$ — not $0.25$.

1. The start of year 5 is four multiplications after purchase: 480,000 → 360,000 → 270,000 → 202,500 → 151,875. The jeepney is worth ₱151,875.
2. Total value lost $= 480{,}000 - 151{,}875 = 328{,}125$, so ₱328,125. Cross-check with the yearly losses: $120{,}000 + 90{,}000 + 67{,}500 + 50{,}625 = 328{,}125$. ✓

(b) Daily collections are geometric with $a_1 = 64{,}000$ and $r = \frac34$.

1. First three days: $64{,}000 + 48{,}000 + 36{,}000 = 148{,}000$, i.e. ₱148,000. The formula agrees: $S_3 = \frac{64{,}000\left(1 - \frac{27}{64}\right)}{1 - \frac34} = 64{,}000 \cdot \frac{37}{64} \cdot 4 = 148{,}000$.
2. Since $\lvert r \rvert = 0.75 < 1$, the infinite total converges: $S_\infty = \frac{64{,}000}{1 - 0.75} = \frac{64{,}000}{0.25} = 256{,}000$. The drive can creep toward ₱256,000 but can never reach it.

8. Common traps

• Off-by-one on $n-1$. The 10th term takes nine steps, not ten. Writing $a_1 + nd$ is the single most frequent arithmetic-sequence error.
• "In the 10th year" versus "over 10 years." The first asks for $a_{10}$, the second for $S_{10}$. Read the preposition before you pick a formula.
• Percent language hides a ratio. "Increases by 6%" means $r = 1.06$; "falls by 6%" means $r = 0.94$. Using $r = 0.06$ turns steady growth into instant collapse.
• Divergent series. When $\lvert r \rvert \ge 1$ there is no infinite sum, however inviting the answer choices look.
• Bouncing objects. The first drop is travelled once; every rebound height is travelled twice — up and then down again.
• Year 0 versus year 1. Decide whether the purchase price is $a_1$ or the value after one year, then keep that choice for the whole item.