8.2 Arithmetic & Geometric Sequences

Official Guide Math Review 3.4 groups series with sets, counting, and estimation. On GMAT Quantitative Reasoning you will not be asked to derive closed forms from first principles. You will be asked to recognize a constant difference, a constant ratio, or a repeating cycle, then apply a short formula by hand in 45 minutes with no calculator.

Arithmetic Sequences

An arithmetic sequence has a constant difference $d = a_{n+1} - a_n$. Once $a_1$ and $d$ are known, every later term is determined:

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

Relating two arbitrary terms: $a_m = a_k + (m - k)d$. The number of terms from $a_1$ through $a_n$ inclusive is

$n = \frac{a_n - a_1}{d} + 1$

The extra $+1$ is the inclusive-count correction. Forgetting it is the most common arithmetic-sequence error on Problem Solving items that ask how many multiples of $k$ lie between two endpoints.

The sum of $n$ arithmetic terms is $n$ times the average of the first and last terms:

$S_n = n \times \frac{a_1 + a_n}{2} = \frac{n}{2}\bigl[2a_1 + (n - 1)d\bigr]$

That average is also the mean of the entire list, and it is the median when $n$ is odd (the middle term) or the average of the two central terms when $n$ is even.

Evenly Spaced Integers

Consecutive integers, consecutive evens, consecutive odds, and the multiples of a fixed integer $k$ are all arithmetic. High-yield special cases:

• First $n$ positive integers: $1 + 2 + \cdots + n = n(n + 1)/2$
• First $n$ odd positives: $1 + 3 + \cdots + (2n - 1) = n^2$
• First $n$ even positives: $2 + 4 + \cdots + 2n = n(n + 1)$
• Multiples of $k$ from $a$ through $b$ inclusive: first = the least multiple of $k$ that is $\ge a$, last = the greatest multiple of $k$ that is $\le b$, then apply the arithmetic-sum formula

Because the list is evenly spaced, the mean equals $(\text{first} + \text{last})/2$. If a GMAT stem gives the mean of an evenly spaced set, you already know the average of the endpoints.

Geometric Sequences

A geometric sequence has a constant ratio $r = a_{n+1}/a_n$ with $r \neq 0$. The $n$-th term is

$a_n = a_1 \cdot r^{n-1}$

The sum of the first $n$ terms, for $r \neq 1$, is

$S_n = a_1 \frac{r^n - 1}{r - 1} = a_1 \frac{1 - r^n}{1 - r}$

If $r = 1$, the sequence is constant and $S_n = n a_1$. GMAT items almost never require a logarithm to recover $n$. Typical common ratios are $2$, $3$, $1/2$, $1/3$, or $10$, so you multiply term-by-term or evaluate a small power such as $2^6 = 64$ by doubling.

Three consecutive positive terms $a$, $b$, $c$ form a geometric sequence if and only if $b^2 = ac$ (the middle term is the geometric mean).

Infinite Geometric Series

If $|r| < 1$, the infinite sum converges:

$S_{\infty} = \frac{a_1}{1 - r}$

If $|r| \ge 1$, the series diverges. A classic application is a bouncing-ball total distance. A ball dropped from height $H$ that rebounds each time to a fraction $r$ of the previous height travels

$H + 2\sum_{k=1}^{\infty} H r^k = H + \frac{2Hr}{1 - r} = H\frac{1 + r}{1 - r}$

The first drop is counted once; every later descent is paired with the preceding ascent. Repeating decimals are the same series: $0.\overline{36} = (36/100)/(1 - 1/100) = 36/99$.

Recursive and Periodic Sequences

A recursive definition such as $a_n = a_{n-1} - a_{n-2}$ does not come with a closed form on the GMAT. Compute the first $5$–$8$ terms by hand and look for a repeating block of length $p$. Once $a_{n+p} = a_n$ for all computed $n$, the $N$-th term is the term in position $R = N \bmod p$ of the block, using position $p$ when the remainder is $0$. The sum of the first $N$ terms is

$S_N = q \times (\text{sum of one cycle}) + (\text{sum of the first } R \text{ terms})$

where $N = qp + R$. If one cycle sums to $0$, most of the work collapses to the leftover terms.

Master Formula Table

Worked Example: Inclusive Multiples

Sum every integer from $200$ through $500$ inclusive that is divisible by $7$.

The first eligible term is $7 \times 29 = 203$. The last is $7 \times 71 = 497$. Term count $n = (497 - 203)/7 + 1 = 42 + 1 = 43$. Sum $= 43 \times (203 + 497)/2 = 43 \times 350 = 15{,}050$.

Worked Example: Finite Geometric Sum

A geometric sequence has first term $5$ and common ratio $2$. The sum of the first six terms is $5(2^6 - 1)/(2 - 1) = 5(64 - 1) = 315$. Multiplying through the list confirms the same total: $5 + 10 + 20 + 40 + 80 + 160 = 315$.

Worked Example: Periodic Recursion

Let $a_1 = 3$, $a_2 = 5$, and $a_n = a_{n-1} - a_{n-2}$ for $n \ge 3$. The terms are $3, 5, 2, -3, -5, -2$, then $3, 5, \ldots$ — a cycle of length $6$ whose sum is $0$. The sum of the first $100$ terms is the leftover after $16$ full cycles ($96$ terms): $a_1 + a_2 + a_3 + a_4 = 3 + 5 + 2 + (-3) = 7$.