This is an arithmetic progression with first term \(a = 4\) and common difference \(d = 13 - 4 = 9\). The nth term is given by \(a_n = a + (n-1)d = 4 + (n-1)9 = 130\). Solving: \(9(n-1) = 126\), so \(n-1 = 14\), giving \(n = 15\). Wait, let me recalculate: \(4 + 9(n-1) = 130\) gives \(9n - 9 = 126\), so \(9n = 135\) and \(n = 15\). Hmm, this doesn't match options either. Actually, checking the pattern: Week 10 would be \(4 + 9(9) = 4 + 81 = 85\), not 130. The answer appears to be beyond the given options.