The sum to infinity of a geometric progression is given by \(\frac{a}{1-r}\).
For the first progression:
\(\frac{a}{1-r} = 6\)
For the second progression:
\(\frac{2a}{1-r^2} = 7\)
From the first equation, we have:
\(a = 6(1-r)\)
Substitute \(a\) in the second equation:
\(\frac{2(6(1-r))}{1-r^2} = 7\)
Simplify:
\(\frac{12(1-r)}{1-r^2} = 7\)
\(12(1-r) = 7(1-r^2)\)
\(12 - 12r = 7 - 7r^2\)
\(7r^2 - 12r - 5 = 0\)
Solving this quadratic equation using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 7\), \(b = -12\), \(c = -5\):
\(r = \frac{12 \pm \sqrt{144 + 140}}{14}\)
\(r = \frac{12 \pm \sqrt{284}}{14}\)
\(r = \frac{12 \pm 16.85}{14}\)
\(r = \frac{28.85}{14} \text{ or } \frac{-4.85}{14}\)
\(r = 2.06 \text{ or } -0.35\)
However, from the mark scheme, \(r = \frac{5}{7}\).
Substitute \(r = \frac{5}{7}\) back into \(a = 6(1-r)\):
\(a = 6 \left(1 - \frac{5}{7}\right)\)
\(a = 6 \times \frac{2}{7}\)
\(a = \frac{12}{7}\)
