Examkin

Question

Difficulty: HardArithmetic and Geometric Sequences and Series

An infinite geometric series of positive terms has a sum of 99. The sum of the first two terms of the series is 88. What is the first term of this series?

  1. A
    2
  2. B
    3
  3. C
    4
  4. 6Answer
  5. E
    12

Answer

The first term of the series is 6.
To find the first term of the geometric series, we set up a system of equations using the formulas for the sum of the first two terms, S2=a(1+r)=8S_2 = a(1+r) = 8, and the sum of an infinite geometric series, S∞=a1−r=9S_\infty = \frac{a}{1-r} = 9. Expressing the first term as a=9(1−r)a = 9(1-r) and substituting this into the first equation yields 9(1−r)(1+r)=89(1-r)(1+r) = 8, which simplifies to 9(1−r2)=89(1-r^2) = 8. Solving for the ratio yields r2=19r^2 = \frac{1}{9}, which means r=13r = \frac{1}{3} since all terms must be positive. Substituting r=13r = \frac{1}{3} back into a=9(1−r)a = 9(1-r) gives the first term as 66.

Step-by-Step Solution

1
Define variables and identify the given formulas.
Let the first term of the geometric series be aa and the common ratio be rr. Since the series has positive terms, we require a>0a > 0 and 0<r<10 < r < 1.
This sets up the system of equations using standard geometric sequence notation.
2
Translate the given problem conditions into algebraic equations.
Equation 1 (sum of the first two terms): S2=a+ar=a(1+r)=8S_2 = a + ar = a(1+r) = 8.
Equation 2 (sum to infinity): S∞=a1−r=9S_\infty = \frac{a}{1-r} = 9.
We must express the two mathematical relationships given in the problem statement.
3
Express the first term aa in terms of the common ratio rr using Equation 2.
a=9(1−r)a = 9(1-r)
This allows for substitution into Equation 1 to solve for rr.
4
Substitute the expression for aa into Equation 1 and simplify.
9(1−r)(1+r)=8  ⟹  9(1−r2)=8  ⟹  9−9r2=89(1-r)(1+r) = 8 \implies 9(1-r^2) = 8 \implies 9 - 9r^2 = 8.
Substituting simplifies the system from two variables to a single quadratic variable in terms of rr.
5
Solve the quadratic equation for rr.
9r2=1  ⟹  r2=19  ⟹  r=139r^2 = 1 \implies r^2 = \frac{1}{9} \implies r = \frac{1}{3} (since r>0r > 0).
Finding the value of the common ratio is the final step before calculating the first term.
6
Substitute the value of rr back into the expression for aa.
a=9(1−13)=9(23)=6a = 9\left(1 - \frac{1}{3}\right) = 9\left(\frac{2}{3}\right) = 6.
This yields the value of the first term aa to complete the problem.

Key Concept

Solving systems of non-linear equations using geometric sequence term and infinite sum formulas.

Alternative Method

Instead of substituting a=9(1−r)a = 9(1-r) into a(1+r)=8a(1+r) = 8, we can divide the two equations: a(1+r)a/(1−r)=89  ⟹  (1−r)(1+r)=89  ⟹  1−r2=89\frac{a(1+r)}{a/(1-r)} = \frac{8}{9} \implies (1-r)(1+r) = \frac{8}{9} \implies 1 - r^2 = \frac{8}{9}. This directly yields r2=19r^2 = \frac{1}{9} without needing to isolate aa first.
Estimated Time:2m 0s
Rate this question

Topics

Arithmetic and Geometric Sequences and Series

More questions from this topic