设 Raiders 四节的得分依次为 \(a, ar, ar^2, ar^3\)（递增等比数列，\(r\gt1\)），Wildcats 的得分依次为 \(a, a+d, a+2d, a+3d\)（递增等差数列），两队第一节都得 \(a\)。
将 \(r=\frac{m}{n}\) 写成最简分数，其中 \(m\gt n\)。因为 \(ar^3\) 是整数，所以 \(a\) 是 \(n^3\) 的倍数；令 \(a=A n^3\)。Raiders 的总分为 \[ R=A(n^3+mn^2+m^2n+m^3)\text{。}\]由于 \(m^3\lt R\le100\)，可知 \(m\le4\)。数对 \((4,3)\) 已使括号内的和达到 \(175\)，所以互质数对 \((m,n)\) 只可能是 \((2,1),(3,1),(3,2),(4,1)\)。
相应的 \(\frac{R}{A}\) 分别为 \(15,40,65,85\)。对于 \((4,1)\) 和 \((3,2)\)，上界迫使 \(A=1\)，但由 \(R-1=4a+6d\) 得到的 \(d\) 不是整数。对于 \((3,1)\)，会得到 \(36A-1=6d\)，这在模 \(6\) 意义下不可能。
当 \((m,n)=(2,1)\) 时，\(R=15A\) 且 \(a=A\)，所以 \(11A-1=6d\)。因而 \(A\equiv5\pmod6\)，再由 \(R\le100\) 得 \(A=5\)。此时 \(d=9\)，Raiders 的得分是 \(5,10,20,40\)，Wildcats 的得分是 \(5,14,23,32\)。Raiders 以 \(75\) 比 \(74\) 获胜。
上半场总分为 \((5+10)+(5+14)=34\)。
因此，正确答案是 E。
English
Let the Raiders score \(a, ar, ar^2, ar^3\) (increasing geometric, \(r\gt1\)) and the Wildcats \(a, a+d, a+2d, a+3d\) (increasing arithmetic), tied in the first quarter at \(a.\)
Write \(r=\frac{m}{n}\) in lowest terms, with \(m\gt n.\) Since \(ar^3\) is an integer, \(a\) is divisible by \(n^3;\) put \(a=A n^3.\) The Raiders’ total is \[ R=A(n^3+mn^2+m^2n+m^3). \] Since \(m^3\lt R\le100,\) we have \(m\le4.\) The pair \((4,3)\) already makes the parenthesized sum \(175,\) so the only possible coprime pairs \((m,n)\) are \((2,1),(3,1),(3,2),(4,1).\)
The corresponding base values of \(\frac{R}{A}\) are \(15,40,65,85.\) For \((4,1)\) and \((3,2),\) the bound forces \(A=1,\) but \(R-1=4a+6d\) gives a nonintegral \(d.\) For \((3,1),\) it would give \(36A-1=6d,\) which is impossible modulo \(6.\)
For \((m,n)=(2,1),\) we have \(R=15A\) and \(a=A,\) so \(11A-1=6d.\) Thus \(A\equiv5\pmod6,\) and \(R\le100\) forces \(A=5.\) Then \(d=9,\) giving Raiders scores \(5,10,20,40\) and Wildcats scores \(5,14,23,32.\) The Raiders win \(75\) to \(74.\)
The first-half total is \((5+10)+(5+14)=34.\)
Thus, the correct answer is E.