Let the first term be \(a\) and the common ratio be \(r\).
The first term is \(a\), the second term is \(ar\), and the third term is \(ar^2\).
According to the problem, we have:
\(a + ar = 50\)
\(ar + ar^2 = 30\)
From the first equation, factor out \(a\):
\(a(1 + r) = 50\)
From the second equation, factor out \(ar\):
\(ar(1 + r) = 30\)
Divide the second equation by the first equation:
\(\frac{ar(1 + r)}{a(1 + r)} = \frac{30}{50}\)
\(r = \frac{3}{5}\)
Substitute \(r = \frac{3}{5}\) back into the first equation:
\(a(1 + \frac{3}{5}) = 50\)
\(a(\frac{8}{5}) = 50\)
\(a = \frac{125}{4}\)
The sum to infinity \(S\) is given by:
\(S = \frac{a}{1 - r}\)
\(S = \frac{\frac{125}{4}}{1 - \frac{3}{5}}\)
\(S = \frac{\frac{125}{4}}{\frac{2}{5}}\)
\(S = \frac{125}{4} \times \frac{5}{2}\)
\(S = \frac{625}{8}\)
