Sequence questions give you a rule for building a list of numbers and ask for a term, a sum or a count. On the GMAT Focus they appear in Quant Problem Solving and in Data Sufficiency, and nearly all of them come down to finding a pattern rather than using a formula.
Start by writing out the first four or five terms. Recursive definitions, where each term depends on the one before, become obvious once you see them on paper, and many sequences repeat in cycles. Once you know the cycle length, the 100th term matches the term whose position is the remainder of 100 divided by the cycle length (a remainder of 0 means the last term of the cycle).
Arithmetic sequences add the same amount each time: the n-th term is a + (n − 1)d, and the sum is the number of terms times the average of the first and last terms. That average trick also handles consecutive integers and evenly spaced sets: the mean equals the median. For counting terms, use (last − first) ÷ spacing + 1, and do not forget the +1.
In Data Sufficiency, a rule relating terms is not enough on its own. You usually need one actual term to pin the sequence down, so check whether a statement gives you a starting value or only a relationship.
4 practice questions
Problem SolvingSequencesFractions/Ratios/DecimalsHard
Question 1
For each positive integer $n$, $a_n = \frac{1}{n(n+1)}$. What is the least positive integer $k$ for which $a_1 + a_2 + \cdots + a_k > 0.98$?
A$50$
B$51$
C$98$
D$99$
E$100$
Show answer & explanation
Correct answer: A
Key Concept or Approach
Split each term: $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. The sum telescopes.
Step 2: Need $1 - \frac{1}{k+1} > 0.98$, i.e. $\frac{1}{k+1} < 0.02 = \frac{1}{50}$, so $k + 1 > 50$.
Step 3: Least integer: $k = 50$ (at $k = 49$ the sum is exactly $0.98$, not greater).
Final Answer
Answer: A
Common trap: 51 - writing the partial sum as 1 - 1/k (an off-by-one in the telescoping), which pushes the answer to 51.
Problem SolvingFractions/Ratios/DecimalsSequencesHard
Question 2
What is the sum of the first $50$ digits after the decimal point in the decimal expansion of $\frac{3}{7}$?
A$216$
B$220$
C$222$
D$225$
E$230$
Show answer & explanation
Correct answer: C
Key Concept or Approach
The decimal for $\frac{3}{7}$ repeats in a block of 6 digits. Count full blocks, then add the leftover digits.
Step-by-Step Solution
Step 1: Long division: $\frac{3}{7} = 0.428571428571\ldots$, repeating block $428571$ with digit sum $4 + 2 + 8 + 5 + 7 + 1 = 27$.
Step 2: $50 = 6 \cdot 8 + 2$: eight full blocks give $8 \cdot 27 = 216$.
Step 3: The next 2 digits are $4$ and $2$: total $216 + 6 = 222$.
Final Answer
Answer: C
Common trap: 225 - multiplying the block sum 27 by 50/6 (about 8.33) instead of adding the actual leftover digits 4 and 2.
Data SufficiencySequencesHard
Question 3
In an arithmetic sequence $a_1, a_2, a_3, \ldots$, what is the sum of the first 11 terms?
(1) $a_6 = 7$
(2) $a_2 + a_{10} = 14$
AStatement (1) ALONE is sufficient, but statement (2) ALONE is NOT sufficient.
BStatement (2) ALONE is sufficient, but statement (1) ALONE is NOT sufficient.
CBOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
DEACH statement ALONE is sufficient.
EStatements (1) and (2) TOGETHER are NOT sufficient.
Show answer & explanation
Correct answer: D
Key Concept or Approach
In an arithmetic sequence, the sum of an odd number of consecutive terms equals the number of terms times the middle term: $S_{11} = 11 \cdot a_6$. Terms equally spaced around $a_6$ average to $a_6$.
