Geometric Sequences Problems with Solutions

REVIEW OF GEOMETRIC SEQUENCES

The sequence shown below

Problem 1:
Find the terms a₂, a₃, a₄ and a₅ of a geometric sequence if a₁ = 10 and the common ratio r = - 1.

Solution to Problem 1:

• Use the definition of a geometric sequence
  
  a₂ = a₁ * r = 10 * (-1) = - 10
  
  a₃ = a₂ * r = - 10 * (-1) = 10
  
  a₄ = a₃ * r = 10 * (-1) = - 10
  
  a₅ = a₄ * r = - 10 * (-1) = 10

Problem 2:
Find the 10 th term of a geometric sequence if a₁ = 45 and the common ration r = 0.2.

Solution to Problem 2:

• Use the formula a_n = a₁ * r^n-1 that gives the n th term to find a₁₀ as follows
  
  a₁₀ = a₁ * r^n-1
  
  = 45 * 0.2⁹ = 2.304 * 10^-5

Problem 3:
Find a₂₀ of a geometric sequence if the first few terms of the sequence are given by

Solution to Problem 3:

• We first use the first few terms to find the common ratio
  
  r = a₂ / a₁ = (1/4) / (-1/2) = -1/2
  
  r = a₃ / a₂ = (-1/8) / (1/4) = -1/2
  
  r = a₄ / a₃ = (1/16) / (-1/8) = -1/2
  
  
• The common ration r = -1/2. We now use the formula a_n = a₁ * r^n-1 for the n th term to find a₂₀ as follows.
  
  a₂₀ = a₁ * r^20-1
  
  = (-1/2) * (-1/2)^20-1 = 1 / (20²⁰)

Problem 4:

Given the terms a₁₀ = 3 / 512 and a₁₅ = 3 / 16384 of a geometric sequence, find the exact value of the term a₃₀ of the sequence.

Solution to Problem 4:

• We first use the formula for the n th term to write a₁₀ and a₁₅ as follows
  
  a₁₀ = a₁ * r^10-1 = 3 / 512
  
  a₁₅ = a₁ * r^15-1 = 3 / 16384
  
  
• We now divide the terms a₁₀ and a₁₅ to write
  
  a₁₅ / a₁₀ = (a₁ * r¹⁴ / a₁ * r⁹) = (3 / 16384) / (3 / 512)
  
  
• Solve for r to obtain.
  
  r⁵ = 1 / 32 which gives r = 1/2
  
  
• We now use a₁₀ to find a₁ as follows.
  
  a₁₀ = 3 / 512 = a₁(1/2)⁹
  
  
• Solve for a₁ to obtain.
  
  a₁ = 3
  
  
• We now use the formula for the n th term to find a₃₀ as follows.
  
  a₃₀ = 3(1/2)²⁹ = 3 / 536870912

Problem 5:
Find the sum

Solution to Problem 5:

• We first rewrite the sum S as follows
  
  S = 1 + 3 + 9 + 27 + 81 + 243 = 364
  
  
• Another method is to first note that the terms making the sum are those of an arithmetic sequence with a₁ = 1 and r = 3 using the formula s_n = a₁ (1 - rⁿ) / (1 - r) with n = 6.
  
  s₆ = 1 (1 - 3⁶) / (1 - 3) = 364

Problem 6:
Find the sum

Solution to Problem 6:

• An examination of the terms included in the sum are
  
  8 , 8*((1/4)¹ , 8*((1/4)² , ... , 8*((1/4)⁹
  
  
• These are the terms of a geometric sequence with a₁ = 8 and r = 1/4 and therefore we can use the formula for the sum of the terms of a geometric sequence
  
  s₁₀ = a₁ (1 - rⁿ) / (1 - r)
  
  = 8 * (1 - (1/4)¹⁰) / (1 - 1/4) = 10.67 (rounded to 2 decimal places)

Problem 7:
Write the rational number 5.31313131... as the ratio of two integers.

Solution to Problem 7:

• We first write the given rational number as an infinite sum as follows
  
  5.313131... = 5 + 0.31 + 0.0031 + 0.000031 + ....
  
  
• The terms making 0.31 + 0.0031 + 0.000031 ... are those of a geometric sequence with a₁ = 0.31 and r = 0.01. Hence the use of the formula for an infinite sum of a geometric sequence
  
  S = a₁ / (1 - r) = 0.31 / (1 - 0.01) = 0.31 / 0.99 = 31 / 99
  
  
• We now write 5.313131... as follows
  
  5.313131... = 5 + 31/99 = 526 / 99