How to use this page
- Work in order. Start at Level 1. Only move up when the method feels automatic.
- Write full working on paper first - never scroll to the answer first.
- Click "Show Answer" to reveal the worked solution and compare your method step by step.
- Note your errors. If you miss a question, write the exact mistake - not just the right answer.
- Repeat the hardest cards. Mastery comes from doing, not reading.
√ 1st differences: \(4;\;8;\;12\)
√ 2nd difference = \(4\)
√ Next 1st differences: \(16;\;20\)
√ \(T_5 = 29 + 16 = 45\)
√ \(T_6 = 45 + 20 = 65\)
√ 1st differences: \(7;\;9;\;11\)
√ 2nd difference = \(2\)
√ Next 1st difference: \(13\)
√ \(T_5 = 40 + 13 = 53\)
√ \(a = 8\)
√ \(d = 7\)
√ \(T_{34} = 8 + (34-1)(7) = 239\)
√ \(r = \dfrac{12}{6} = 2\)
√ Previous term: \(\dfrac{6}{2} = 3\)
√ \(S_{20} = \dfrac{20}{2}\bigl[2(7) + 19(5)\bigr]\)
√ \(S_{20} = 10(14 + 95)\)
√ \(S_{20} = 10(109) = 1090\)
√ \(S_\infty = \dfrac{a}{1-r}\) only valid for \(|r| < 1\)
√ \(|r| = 3 \geq 1\)
√ Series diverges
√ \(S_\infty\) does not exist
√ 1st differences: \(5;\;9;\;13\)
√ \(T_4 = 16 + 13 = 29\)
√ \(T_n = 2n^2 - n + 1\)
√ \(2n^2 - n + 1 = 862\)
√ \(2n^2 - n - 861 = 0\)
√ \(n = 21\)
√ 21 terms
√ 1st differences: \(10;\;14;\;18\)
√ 2nd difference = \(4\)
√ Next 1st differences: \(22;\;26\)
√ \(T_5 = 51 + 22 = 73\)
√ \(T_6 = 73 + 26 = 99\)
√ 2nd difference = \(4\)
√ \(a = \frac{4}{2} = 2\)
√ \(T_n = 2n^2 + bn + c\)
√ \(2 + b + c = 5 \Rightarrow b + c = 3\)
√ \(8 + 2b + c = 9 \Rightarrow 2b + c = 1\)
√ \(b = -2\)
√ \(c = 5\)
√ \(T_n = 2n^2 - 2n + 5\)
√ 2nd difference = \(2\)
√ \(a = 1\)
√ \(b = 4\)
√ \(c = 8\)
√ \(T_n = n^2 + 4n + 8\)
√ \(n^2 + 4n + 8 = 4493\)
√ \(n^2 + 4n - 4485 = 0\)
√ \(n = \dfrac{-4 + \sqrt{16 + 17940}}{2} = 65\)
√ \(n = 65 \in \mathbb{N}\)
√ 4493 is the 65th term
√ \((2p-3)-(1-p) = (p+5)-(2p-3)\)
√ \(3p - 4 = -p + 8\)
√ \(4p = 12\)
√ \(p = 3\)
√ \(a = 1 - 3 = -2\)
√ \(d = 2(3)-3-(-2) = 5\)
√ 1st differences: \(7;\;5\)
√ 2nd difference = \(-2\)
√ \(a = -1\)
√ \(9a + 3b + c = 0\)
√ \(a + b + c = T_1\)
√ \(b = 14\)
√ \(c = -33\)
√ \(T_n = -n^2 + 14n - 33\)
√ \(T_1 = -1 + 14 - 33 = -20\)
√ 1st differences: \(4-m;\;n-4;\;22-n\)
√ 2nd difference = \(4\)
√ \((n-4)-(4-m) = 4 \Rightarrow n + m = 12\)
√ \((22-n)-(n-4) = 4 \Rightarrow n = 11\)
√ \(m = 1\)
√ \(a = 2\)
√ \(T_n = 2n^2 - 3n + 2\)
√ 1st differences: \(12;\;16\)
√ 2nd difference = \(4\)
√ Next 1st differences: \(20;\;24\)
√ \(a = 40 + 20 = 60\)
√ \(b = 60 + 24 = 84\)
√ \(T_n = 2n^2 + 6n + 4\)
√ \(2n^2 + 6n + 4 = 312\)
√ \(2n^2 + 6n - 308 = 0\)
√ \(n^2 + 3n - 154 = 0\)
√ \(n = 11\)
√ Day 11
√ 1st differences: \(14;\;12;\;10\)
√ 2nd difference = \(-2\)
√ \(a = -1\)
√ \(b = 17\)
√ \(c = -1\)
√ \(T_n = -n^2 + 17n - 1\)
√ \(T_5 = -25 + 85 - 1 = 59\)
√ \(T_{27} = -729 + 459 - 1 = -271\)
√ 1st differences: \(2;\;6;\;10;\;14\)
√ \(T_1 = 8\)
√ \(T_2 = 8 + 2 = 10\)
√ \(T_3 = 10 + 6 = 16\)
√ 2nd difference = \(4\)
√ \(a = 2\)
√ \(b = -4\)
√ \(c = 10\)
√ \(T_n = 2n^2 - 4n + 10\)
√ \(4x - (x+2) = (6x+4) - 4x\)
√ \(3x - 2 = 2x + 4\)
√ \(x = 6\)
√ Sequence: \(8;\;24;\;40\)
√ \(d = 16\)
√ 2nd difference = \(2\)
√ \(a = 1\)
√ \(b = 3\)
√ \(c = 1\)
√ \(T_n = n^2 + 3n + 1\)
√ \(n^2 + 3n + 1 = 109\)
√ \(n^2 + 3n - 108 = 0\)
√ \((n-9)(n+12) = 0\)
√ \(n = 9\)
(1) Calculate the sum of the first 25 even-positioned terms.
(2) Use \(4, 13, 22, \ldots\) as the first differences of a new sequence whose first term is \(-6\). Find the new general term.
√ (1) Even terms: \(a = 13,\ d = 18,\ n = 25\)
√ \(S_{25} = \dfrac{25}{2}\bigl[2(13) + 24(18)\bigr]\)
√ \(S_{25} = \dfrac{25}{2}(458) = 5725\)
√ (2) \(T_1 = -6\)
√ \(T_n = -6 + \displaystyle\sum_{k=1}^{n-1}(9k-5)\)
√ \(T_n = \dfrac{9n^2 - 19n - 2}{2}\)
√ \(a = 1,\ r = 3\)
√ Sum \(= \dfrac{3^n - 1}{2}\)
√ \(\dfrac{5}{3} \cdot \dfrac{3^n - 1}{2} = \dfrac{1820}{3}\)
√ \(5(3^n - 1) = 3640\)
√ \(3^n - 1 = 728\)
√ \(3^n = 729 = 3^6\)
√ \(n = 6\)
√ \(a = 8\)
√ \(d = 7\)
√ \(S_{40} = \dfrac{40}{2}\bigl[2(8) + 39(7)\bigr]\)
√ \(S_{40} = 20(16 + 273)\)
√ \(S_{40} = 20(289) = 5780\)
√ \(a = 5,\ r = 0.6\)
√ \(|r| < 1\)
√ Series converges
√ \(S_\infty = \dfrac{5}{1 - 0.6} = 12.5\)
√ \(12.5 < 13\)
√ Ball will never reach the hole
√ \(r = \frac{1}{2}\)
√ \(|r| < 1\)
√ Series converges
√ \(S_\infty = \dfrac{10}{1 - \frac{1}{2}} = 20\)
√ \(S_n = \dfrac{10\left(1 - \left(\frac{1}{2}\right)^n\right)}{1 - \frac{1}{2}}\)
√ \(S_n = 20\left(1 - \dfrac{1}{2^n}\right)\)
√ \(S_\infty - S_n = 20 - 20\left(1 - \dfrac{1}{2^n}\right)\)
√ \(S_\infty - S_n = 20 \cdot \left(\dfrac{1}{2}\right)^n\)
√ \(T_{n+1} - T_n = 2(n+1)^2 - 2(n+1) + 3 - (2n^2 - 2n + 3)\)
√ \(d_n = 4n\)
√ \[\displaystyle\sum_{k=1}^{50} 4k\]
(1) Determine \(T_5\). (2) Find the general term. (3) If the sum of the first 40 first differences is \(3320\), identify which term equals \(3322\).
√ (1) 1st differences: \(5;\;9;\;13\)
√ Next 1st difference: \(17\)
√ \(T_5 = 29 + 17 = 46\)
√ (2) 2nd difference = \(4\)
√ \(a = 2\)
√ \(T_n = 2n^2 - n + 1\)
√ (3) \(\displaystyle\sum_{k=1}^{40} d_k = T_{41} - T_1\)
√ \(T_{41} = 2 + 3320 = 3322\)
√ \(T_{41} = 3322\)
√ Numerator: \(2n + 1\)
√ Denominator: \(2^{n-1}\)
√ \[\displaystyle\sum_{n=1}^{k} \dfrac{2n+1}{2^{n-1}}\]
√ \(d_m = 2m - 1\)
√ \(d_{98} = 2(98) - 1 = 195\)
√ \(T_{98} = 9632 - 195 = 9437\)
√ 2nd difference = \(2\)
√ \(a = 1\)
√ \(T_n = n^2 + bn + c\)
√ \(9 + 3b + c = 32 \Rightarrow 3b + c = 23\)
√ \(T_1 = 32 - 3 - 1 = 28\)
√ \(1 + b + c = 28 \Rightarrow b + c = 27\)
√ \(b = -2\)
√ \(c = 29\)
√ \(T_n = n^2 - 2n + 29\)
√ 1st differences: \(1;\;-1;\;-3;\;-5\)
√ \(a_d = 1,\ d_d = -2\)
√ \(d_n = 1 + (n-1)(-2) = 3 - 2n\)
√ 2nd difference = \(-2\)
√ \(a = -1\)
√ \(b = 4\)
√ \(c = -6\)
√ \(T_n = -n^2 + 4n - 6\)
√ \(T_n = -(n-2)^2 - 2\)
√ \(-(n-2)^2 \leq 0\)
√ \(T_n \leq -2 < 0\)
√ Sequence never contains a positive term
√ 1st differences: \(-3;\;-5;\;-7\)
√ \(a_d = -3,\ d_d = -2\)
√ 2nd difference = \(-2\)
√ \(a = -1\)
√ \(b = 0\)
√ \(c = 12\)
√ \(T_n = 12 - n^2\)
√ \(d_m = -2m - 1\)
√ \(-2m - 1 = -115\)
√ \(m = 57\)
√ Between \(T_{57}\) and \(T_{58}\)
√ \(2n^2 + n - 9 = 519\)
√ \(2n^2 + n - 528 = 0\)
√ \(n = \dfrac{-1 + \sqrt{1 + 4224}}{4}\)
√ \(n = \dfrac{-1 + 65}{4} = 16\)
√ \(T_{16} = 519\)
√ \(2n^2 - 2n + 3 = 4903\)
√ \(2n^2 - 2n - 4900 = 0\)
√ \(n^2 - n - 2450 = 0\)
√ \(n = \dfrac{1 + \sqrt{1 + 9800}}{2} = 50\)
√ \(T_{50} = 4903\)
√ \(d_m = 4m\)
√ \(4m = 2000\)
√ \(m = 500\)
√ Between \(T_{500}\) and \(T_{501}\)
√ 1st differences: \(2x-5;\;x+2;\;x+4\)
√ \((x+2)-(2x-5) = (x+4)-(x+2)\)
√ \(-x + 7 = 2\)
√ \(x = 5\)
√ 2nd difference = \(2\)
√ \(a = 1\)
√ \(b = 2\)
√ \(c = 2\)
√ \(T_n = n^2 + 2n + 2\)
√ \(T_n = (n+1)^2 + 1\)
√ \((n+1)^2 + 1 \geq 2 > 0\)
√ All terms are positive
√ \(k = 2\): first term \(= 6\)
√ \(k = n\): last term \(= 2(3^{n-1})\)
√ \(a = 6,\ r = 3\)
√ Number of terms \(= n-1\)
√ \(S = \dfrac{6(3^{n-1} - 1)}{2} = 3(3^{n-1} - 1)\)
√ \(3(3^{n-1} - 1) = 59\,046\)
√ \(3^{n-1} - 1 = 19\,682\)
√ \(3^{n-1} = 19\,683 = 3^9\)
√ \(n - 1 = 9\)
√ \(n = 10\)
√ \(T_n = S_n - S_{n-1}\)
√ \(T_n = (2n - n^2) - (2(n-1)-(n-1)^2)\)
√ \(T_n = 3 - 2n\)
√ \(T_1 = 3 - 2 = 1\)
√ \(T_{13} = -23\)
√ \(T_{14} = -25\)
√ \(T_{15} = -27\)
√ Sum \(= -23 + (-25) + (-27) = -75\)
What The Examiner Is Really Testing
Method Over Answer
In Patterns & Sequences, marks are awarded for method steps, not just the final answer. Always show: (1) identify the type, (2) write the formula, (3) substitute carefully, (4) verify. Even if you slip on arithmetic, method marks are still available.
The Role of Convergence
Every time you use \(S_\infty\), you must state \(|r| < 1\) and verify it. The examiner awards a mark specifically for this check. Skipping it costs you marks even when the rest of the calculation is correct.
Connecting Levels
L3 and L4 questions often link concepts - e.g., using a sum rule to recover individual terms, or using first differences to identify between which consecutive terms a condition holds. Practise moving between these representations.
Justification Earns Marks
Questions that ask "explain why" or "show that" require a mathematical reason, not just a statement. Completing the square, using range arguments, or citing the convergence condition are all examinable forms of justification.