① 与えられた等比数列を {\(a_{n}\)} 公比を r (∈R) とする。
第 n 部分和を \(S_{n}\) と置くと, 題意より明らかに r ≠ 1 で
\(S_{10}\) = \(a_{1}\)(\(r^{10}\) - 1)/(r - 1) = 4.
\(S_{20}\) - \(S_{10}\) = \(a_{1}\)・\(r^{10}\)(\(r^{20}\) - 1)/(r - 1) = 24
(初項 \(a_{11}\), 公比 r の等比数列の 20 項の和だから).
下の式を上の式で割る。
\(r^{10}\)・(\(r^{20}\) - 1)/(\(r^{10}\) - 1) = 6
\(r^{10}\)・(\(r^{10}\) + 1) = 6
(\(r^{10}\)\()^{2}\) + \(r^{10}\) - 6 = (\(r^{10}\) - 2)(\(r^{10}\) + 3) = 0.
\(r^{10}\) = 2, -3 だが r ∈R より \(r^{10}\) = (\(r^{2}\)\()^{5}\) > 0 だから
\(r^{10}\) = 2.
従って, 求める \(S_{30}\) - \(S_{20}\) は初項 \(a_{21}\), 公比 r,
項数が 30 の和だから
\(a_{1}\)・\(r^{20}\)(\(r^{30}\) - 1)/(r - 1)
= (\(r^{10}\)\()^{2}\)・\(a_{1}\)(\(r^{10}\) - 1)(\(r^{20}\) + \(r^{10}\) + 1)/(r - 1)
= 4・[\(a_{1}\)(\(r^{10}\) - 1)/(r - 1)]・((\(r^{10}\)\()^{2}\) + \(r^{10}\) + 1)
= 4×4×(4 + 2 + 1) = 16×7 = 112.

② 題意より
5(2a + 4b)/2 = 10 (⇔a + 2b = 2⇔a = 2 - 2b)
b = 1 の時は a = 0 だから等比数列の和も 0 になるので
b ≠1. 従って題意より
a(\(b^{5}\) - 1)/(b - 1) = \(\frac{31}{16}\).

代入して
2(1-b)(\(b^{5}\) - 1)/(b - 1) = \(\frac{31}{16}\).
-2(\(b^{5}\) - 1) = \(\frac{31}{16}\)
32(\(b^{5}\) - 1) = -31
32 \(b^{5}\) - 32 = -31
32 \(b^{5}\) = 1
\(b^{5}\) = \(\frac{1}{32}\) = 1/\(2^{5}\).
故に b = \(\frac{1}{2}\).
a = 2 - 2b = 2 - 1 = 1.