Sequences and Series — GMAT Focus Questions

27 GMAT Focus practice questions on Sequences and Series, part of Quantitative Reasoning. Below are 12 of them in full, each with the answer and a written explanation.

Questions & explanations

1. The sum of the first 10 terms of an arithmetic series is 200, and the sum of the first 20 terms is 800. What is the sum of the first 30 terms?

  1. 1200
  2. 1400
  3. 1600
  4. 1800
  5. 2000

Answer: 1800

Let first term a, common difference d. Sum of first n terms: S_n = n/2 * (2a + (n-1)d). For n=10: 5*(2a+9d)=200 => 2a+9d=40. For n=20: 10*(2a+19d)=800 => 2a+19d=80. Subtract: 10d=40 => d=4. Then 2a+9*4=40 => 2a=4 => a=2. Then S_30 = 30/2*(2*2+29*4)=15*(4+116)=15*120=1800.

2. A bank offers a savings account with an annual interest rate of 6%, compounded quarterly. If $5,000 is deposited initially and no further deposits or withdrawals are made, what is the total amount in the account after 3 years? (Round to the nearest cent.)

  1. $5,955.08
  2. $5,970.26
  3. $5,978.09
  4. $6,014.52
  5. $6,050.00

Answer: $5,978.09

The amount after t years with quarterly compounding is A = P(1 + r/n)^(nt). Here P = 5000, r = 0.06, n = 4, t = 3. So A = 5000(1 + 0.06/4)^(12) = 5000(1.015)^12. Compute (1.015)^12 ≈ 1.195618, then A ≈ 5000 * 1.195618 = 5978.09. Thus the amount is $5,978.09.

3. In a geometric series, the sum of the first 3 terms is 21 and the sum of the first 6 terms is 189. What is the common ratio?

  1. 1
  2. 2
  3. 3
  4. 4
  5. 5

Answer: 2

Let first term a, common ratio r (r ≠ 1). S_3 = a(1-r^3)/(1-r) = 21, S_6 = a(1-r^6)/(1-r) = 189. Divide S_6 by S_3: (1-r^6)/(1-r^3) = 189/21 = 9. Factor numerator: 1-r^6 = (1-r^3)(1+r^3). So (1-r^3)(1+r^3)/(1-r^3) = 1+r^3 = 9 => r^3 = 8 => r = 2.

4. A sequence is defined recursively by a_1 = 4 and a_n = 3a_{n-1} - 2 for n ≥ 2. Which of the following is an explicit formula for a_n?

  1. a_n = 3^n + 1
  2. a_n = 3^{n-1} + 1
  3. a_n = 3^n - 1
  4. a_n = 3^{n-1} - 1
  5. a_n = 2 * 3^{n-1} + 2

Answer: a_n = 3^n + 1

Compute terms: a1=4, a2=3*4-2=10, a3=3*10-2=28. Check options: a) 3^1+1=4, 3^2+1=10, 3^3+1=28. Works. Alternatively, solve recurrence: a_n - 1 = 3(a_{n-1} - 1) => a_n - 1 = 3^{n-1}(a_1-1)=3^{n-1}*3=3^n => a_n=3^n+1.

5. The repeating decimal 0.\overline{72} can be expressed as the fraction \(\frac{8}{11}\). What is the sum of the infinite geometric series represented by the repeating decimal 0.\overline{324}?

  1. \(\frac{36}{111}\)
  2. \(\frac{12}{37}\)
  3. \(\frac{324}{999}\)
  4. \(\frac{108}{333}\)
  5. \(\frac{4}{11}\)

Answer: \(\frac{12}{37}\)

0.\overline{324} = 0.324324324... = 324/999 = simplify dividing by 27: 12/37. Alternatively, sum of infinite geometric series: first term = 324/1000, r = 1/1000, sum = (324/1000)/(1 - 1/1000) = 324/999 = 12/37.

6. Consider the sequence: 2, 5, 10, 17, 26, ... What is the next term?

  1. 35
  2. 37
  3. 39
  4. 41
  5. 43

Answer: 37

Pattern: differences: 3,5,7,9,... increasing by 2 each time. Next difference = 11, so next term = 26 + 11 = 37. Alternatively, formula: n^2 + 1: 1^2+1=2, 2^2+1=5, 3^2+1=10, 4^2+1=17, 5^2+1=26, 6^2+1=37.

7. What is the sum of the first 10 terms of the alternating series: 3 - 6 + 12 - 24 + ... ?

  1. -1023
  2. -1025
  3. 1023
  4. 1025
  5. 2047

Answer: -1023

This is a geometric series with first term a=3, common ratio r=-2. Sum of first n terms: S_n = a(1 - r^n)/(1 - r). For n=10: S_10 = 3(1 - (-2)^10)/(1 - (-2)) = 3(1 - 1024)/(3) = 1 - 1024 = -1023.

8. Consider the alternating sequence: -2, 4, -8, 16, -32, ... What is the 10th term?

  1. 1024
  2. -1024
  3. 2048
  4. -2048
  5. 512

Answer: 1024

Sequence: a_n = (-1)^n * 2^n? Check: n=1: (-1)^1*2^1=-2; n=2: (-1)^2*2^2=4; n=3: (-1)^3*2^3=-8. So a_n = (-2)^n? (-2)^1=-2, (-2)^2=4, (-2)^3=-8, so a_n = (-2)^n. For n=10, (-2)^10 = 1024.

9. The sum of the first 20 terms of an arithmetic sequence is 610. If the first term is 5, what is the common difference?

  1. 2
  2. 3
  3. 4
  4. 5
  5. 6

Answer: 3

Sum of first n terms: S_n = n/2 * [2a + (n-1)d]. Here n=20, S_20=610, a=5. So 610 = 20/2 * [10 + 19d] = 10*(10+19d) => 61 = 10+19d => 19d=51 => d=51/19=3. Thus common difference is 3.

10. In an arithmetic sequence, the 5th term is 17 and the 12th term is 38. Which term of the sequence is equal to 53?

  1. 15th
  2. 16th
  3. 17th
  4. 18th
  5. 19th

Answer: 17th

Let first term a, common difference d. a5 = a+4d = 17, a12 = a+11d = 38. Subtract: 7d=21 => d=3. Then a=17-4*3=5. nth term: a_n = 5 + (n-1)*3 = 53 => 3(n-1)=48 => n-1=16 => n=17.

11. Which of the following sequences is an arithmetic sequence?

  1. 2, 4, 8, 16, 32
  2. 3, 6, 9, 12, 15
  3. 1, 4, 9, 16, 25
  4. 5, 10, 20, 40, 80
  5. 2, 5, 9, 14, 20

Answer: 3, 6, 9, 12, 15

An arithmetic sequence has a constant common difference between consecutive terms. For option b: 6-3=3, 9-6=3, 12-9=3, 15-12=3. The common difference is 3, so it is arithmetic.

12. The sum of the first n terms of a geometric sequence is 765. The first term is 3 and the common ratio is 2. What is n?

  1. 6
  2. 7
  3. 8
  4. 9
  5. 10

Answer: 8

Sum of first n terms of geometric series: S_n = a(r^n - 1)/(r-1) for r>1. Here a=3, r=2, S_n=765. So 765 = 3(2^n - 1)/(2-1) = 3(2^n - 1) => 255 = 2^n - 1 => 2^n = 256 => n=8.

More Quantitative Reasoning topics

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