Question

The sum of the infinite geometric series $ S = \frac{a}{1-r} $ is 24, and the sum of the first three terms is 21. Find $ a $ and $ r $.

• A. $ a = 12, r = \frac{1}{2} $
• B. $ a = 8, r = \frac{2}{3} $
• C. $ a = 6, r = \frac{3}{4} $
• D. $ a = 10, r = \frac{4}{5} $

Hint:

For an infinite geometric series, use $ S = \frac{a}{1 - r} $. For the sum of the first $ n $ terms, use $ S_n = a \frac{1 - r^n}{1 - r} $.

Answer 1

The Correct Option is A

The problem involves solving for the first term a and the common ratio r of an infinite geometric series given its sum and the sum of the first three terms. The sum of an infinite geometric series is given by the formula: 

\(S = \frac{a}{1-r}\),

where \(S = 24\) according to the problem. Therefore, we have:

\(\frac{a}{1-r} = 24\) ……… (1)

The sum of the first three terms of a geometric series can be expressed as:

\(a + ar + ar^2 = 21\) ……… (2)

We now have two equations to solve for a and r.

From equation (1): \( a = 24(1 - r) \).

Substitute into equation (2):

\( 24(1 - r) + 24(1 - r)r + 24(1 - r)r^2 = 21 \)

Simplifying, we get:

\( 24(1 - r + r - r^2 + r^2 - r^3) = 21 \)

\(24(1 - r^3) = 21\)

Divide both sides by 24:

\(1 - r^3 = \frac{21}{24} = \frac{7}{8}\)

Therefore, \(r^3 = \frac{1}{8}\) which means \(r = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}\).

Substitute back to find a:

\(a = 24(1 - \frac{1}{2}) = 24 \times \frac{1}{2} = 12\).

Thus, a = 12 and r = \(\frac{1}{2}\).

Answer 2

Step 1: Use the formula for the sum of an infinite geometric series.
The sum of the infinite series is given by: $$ S = \frac{a}{1 - r} = 24 $$ This implies: $$ a = 24(1 - r) \quad \text{(Equation 1)} $$ Step 2: Use the sum of the first three terms.
The sum of the first three terms is: $$ a + ar + ar^2 = 18 $$ Factor out $ a $: $$ a(1 + r + r^2) = 18 \quad \text{(Equation 2)} $$ Step 3: Substitute $ a $ from Equation 1 into Equation 2.
From Equation 1: $$ a = 24(1 - r) $$ Substitute this into Equation 2: $$ 24(1 - r)(1 + r + r^2) = 18 $$ Simplify using the identity $ 1 - r^3 = (1 - r)(1 + r + r^2) $: $$ 24(1 - r^3) = 18 $$ Divide both sides by 6: $$ 4(1 - r^3) = 3 $$ Solve for $ r^3 $: $$ 1 - r^3 = \frac{3}{4} \quad \Rightarrow \quad r^3 = \frac{1}{4} $$ Take the cube root: $$ r = \sqrt[3]{\frac{1}{4}} = \frac{1}{\sqrt[3]{4}} \approx \frac{1}{2} $$ Step 4: Solve for $ a $ using $ r $.
From Equation 1: $$ a = 24(1 - r) $$ Substitute $ r = \frac{1}{2} $: $$ a = 24\left(1 - \frac{1}{2}\right) = 24 \cdot \frac{1}{2} = 12 $$ Step 5: Verify the solution.
Sum of the infinite series: $$ S = \frac{a}{1 - r} = \frac{12}{1 - \frac{1}{2}} = \frac{12}{\frac{1}{2}} = 24 \quad \text{(correct)} $$ Sum of the first three terms: $$ a + ar + ar^2 = 12 + 12 \cdot \frac{1}{2} + 12 \cdot \left(\frac{1}{2}\right)^2 = 12 + 6 + 3 = 18 \quad \text{(correct)} $$ Both conditions are satisfied.