P4.4Arithmetic sequences and series

• M1 for \(a + 4d = 29\) and \(a + 13d = 83\)
• M1 for solving simultaneously (e.g. \(9d = 54\))
• A1 for \(d = 6\) and \(a = 5\)

Worked solution: \(a + 4d = 29\) and \(a + 13d = 83\). Subtracting: \(9d = 54\), so \(d = 6\).
Then \(a = 29 - 4 \times 6 = 5\).

• M1 for \(\frac{35}{2}\left(2 \times 5 + 34 \times 6\right)\) with their \(a\) and \(d\)
• A1 for \(3745\)

Worked solution: \(S_{35} = \frac{35}{2}\left(2 \times 5 + 34 \times 6\right) = 3745\)

• M1 for \(-11 + 4(n - 1) \gt 0\) (or listing terms) to find the value of \(n\)
• A1 for \(1\)

Worked solution: \(-11 + 4(n - 1) \gt 0 \Rightarrow n \gt \frac{15}{4} = 3.75\), so the first positive term is the 4th term, \(-11 + 3 \times 4 = 1\).

• M1 for \(\frac{n}{2}\left[2 \times (-11) + (n - 1) \times 4\right] \gt 1000\) (or \(=\))
• A1 for a correct 3-term quadratic, e.g. \(2n^{2} - 13n - 1000 \gt 0\)
• M1 for solving their quadratic to find the positive critical value, e.g. \(n = \frac{13 + \sqrt{8169}}{4}\)
• A1 for \(n = 26\)

Worked solution: \(\frac{n}{2}\left[-22 + 4(n - 1)\right] \gt 1000 \Rightarrow 2n^{2} - 13n - 1000 \gt 0\).
The positive root of \(2n^{2} - 13n - 1000 = 0\) is \(n = \frac{13 + \sqrt{8169}}{4} = 25.85\).
So the least value is \(n = 26\). (Check: \(S_{25} = 925\), \(S_{26} = 1014\).)

• M1 for \(S_n = a + (a + d) + \dots + [a + (n - 1)d]\) and the same series written in reverse
• M1 for adding term by term: each pair sums to \(2a + (n - 1)d\)
• A1* for \(2S_n = n[2a + (n - 1)d]\) and the result

Worked solution: \(S_n = a + (a + d) + \dots + [a + (n - 2)d] + [a + (n - 1)d]\)
\(S_n = [a + (n - 1)d] + [a + (n - 2)d] + \dots + (a + d) + a\)
Adding, each of the \(n\) columns sums to \(2a + (n - 1)d\), so \(2S_n = n[2a + (n - 1)d]\) and \(S_n = \frac{n}{2}[2a + (n - 1)d]\).

• M1 for \(u_n = S_n - S_{n - 1} = 3n^2 + 5n - [3(n - 1)^2 + 5(n - 1)]\) for \(n \ge 2\)
• A1 for \(u_n = 6n + 2\), and checking \(u_1 = S_1 = 8\) fits the formula
• A1* for \(u_{n + 1} - u_n = 6\), constant, so the sequence is arithmetic

Worked solution: For \(n \ge 2\): \(u_n = S_n - S_{n - 1} = 3n^2 + 5n - 3(n - 1)^2 - 5(n - 1) = 3(2n - 1) + 5 = 6n + 2\).
Also \(u_1 = S_1 = 8 = 6(1) + 2\), so \(u_n = 6n + 2\) for all \(n \ge 1\).
\(u_{n + 1} - u_n = 6\), a constant, so the sequence is arithmetic (first term 8, common difference 6).

• M1 for showing that squares leave remainder 0 or 1 on division by 3 (from \((3k)^2\) and \((3k \pm 1)^2 = 3(3k^2 \pm 2k) + 1\))
• M1 for \(6n + 2 = 3(2n) + 2\), remainder 2 on division by 3
• A1* for a conclusion

Worked solution: Any integer is \(3k\) or \(3k \pm 1\). \((3k)^2 = 3(3k^2)\) and \((3k \pm 1)^2 = 3(3k^2 \pm 2k) + 1\), so a square leaves remainder 0 or 1 on division by 3.
But \(u_n = 6n + 2 = 3(2n) + 2\) leaves remainder 2. So no term is a square number.