The problem requires determining the initial term a and the common ratio r of an infinite geometric series, given its sum and the sum of its first three terms. The formula for the sum of an infinite geometric series is:
\(S = \frac{a}{1-r}\).
Given \(S = 24\), we establish the first equation:
\(\frac{a}{1-r} = 24\) ……… (1)
The sum of the first three terms is given by:
\(a + ar + ar^2 = 21\) ……… (2)
We now have a system of two equations with two unknowns.
From equation (1), isolate a: \( a = 24(1 - r) \).
Substitute this expression for a into equation (2):
\( 24(1 - r) + 24(1 - r)r + 24(1 - r)r^2 = 21 \)
Factor out \(24(1 - r)\) and simplify:
\( 24(1 - r)(1 + r + r^2) = 21 \)
Recognize that \( (1 - r)(1 + r + r^2) = 1 - r^3 \):
\(24(1 - r^3) = 21\)
Divide both sides by 24:
\(1 - r^3 = \frac{21}{24} = \frac{7}{8}\)
Solve for \(r^3\): \(r^3 = 1 - \frac{7}{8} = \frac{1}{8}\).
Calculate r: \(r = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}\).
Substitute the value of r back into the equation for a:
\(a = 24(1 - \frac{1}{2}) = 24 \times \frac{1}{2} = 12\).
The solutions are a = 12 and r = \(\frac{1}{2}\).