Number Patterns Sequences and Series Questions and Answers Mathematics Grade 12 Pdf Download

1. 1. It is an arithmetic sequence because there is a common difference.
      a = 6; d = 7 Tn = a + (n – 1)d
      Tn = 6 + (n – 1)(7)
      Tn = 7n – 1
   2. T21 = 7(21) – 1 = 147 – 1 = 146
   3. 97 = 7n – 1
      ∴98 = 7n
      ∴14 = n
      ∴97 is the 14th term of the sequence. (5)
2. 1. It is an arithmetic sequence: a = 8; d = 5 – 8 = 2 – 5 = – 3
      T_n = a + (n – 1)d
      ∴T₁₅ = 8 + (15 – 1)(–3)
      T₁₅ = 8 + 14(–3)
      T₁₅ = 8 – 42 = –34
   2. T_n = a + (n – 1)d
      – 289 = 8 + (n – 1)(–3)
      ∴– 289 = 8 – 3n +3
      ∴– 300 = –3n
      ∴100 = n 3 ∴– 289 will be the 100th term (4)
3. 1. Since this is an arithmetic sequence, you can assume that there is a common difference between the terms.
      d = T₂ – T₁ = T₃ − T₂
      ∴(2p – 3) – (1 – p) = ( p + 5) – (2p – 3)
      3p – 4 = – p + 8
      4p = 12
   2. p = 3
      T₁ = 1 – p = 1 – 3 = –2
      T₂ = 2p – 3 = 2(3) – 3 = 3
      T₃ = p + 5 = 3 + 5 = 8
      So the first three terms of the sequence are –2; 3; 8 (5)
      [14] 

1. 1. It helps to make a diagram:
      
      ∴ it is a quadratic sequence.
      2a = 8 ∴ a = 4
      3a + b = 10 ∴3(4) + b = 10
      b = –2
      a + b + c = 3 ∴ 4 + (–2) + c = 3
      c = 1
      ∴Tn =4n² – 2n + 1
   2. T₇ = 4(7)² – 2(7) + 1
      = 4(49) – 14 + 1
      = 183
   3. 241 = 4n² – 2n + 1
      0 = 4n² – 2n + 1 – 241 make the equation = 0 to solve
      0 = 4n² – 2n – 240
      0 = 2n² – n – 120 divide through by 2
      0 = (2n + 15)(n – 8)
      factorise
      ∴ 2n + 15 = 0 OR n – 8 = 0
      ∴ n = –7,5 OR n = 8 3 (9)
2. T₁       T₂        T₃         T₄         T₅
   –1      3         9           17       27 …
   4       6          8          10
   2         2           2
   ∴T₆ = 27 + (10 + 2) = 39 3 use the pattern of the numbers
   2a = 2 ∴a = 1
   3a + b = 4
   3(1) + b = 4 ∴ b = 1
   a + b + c = –1
   1 + 1 + c = –1 ∴ c = –3
   Tn = n² + n – 3  (4)
   [13]

1. a = 3; r = ⁶/₃ = ¹²/₆ = 2
   Tn = arⁿ⁻¹
   T10 = 3(2)¹⁰⁻¹ = 3(2)⁹ = 3 × 512 = 1536 (2)
2. a = 2; r = ⁴/₂= ⁸/₄ = 2
   ar^n–1 = 1024
   2(2)^n–1 = 2¹⁰ = 2ⁿ = 2¹⁰ 
   ∴ n = 10 (2)
3. ^x/₅ = ⁴⁵/_x
   x = ± √225 = ± 15 (2)
4. ar⁷ = 9
   ar⁹ = 25
   ar⁹ = 25
   ar⁷     9
   ∴r² = ²⁵/₉
   r = ⁵/₃
   
   [9] 

1. a = 19 and d = 3
   T_n = 3n + 16 = 121
   3n = 105
   n = 35
   Sn  = ⁿ/₂(a + l)
   S₃₅  = ³⁵/₂ (19 + 121) =  ³⁵/₂ (140) = 35 × 70 = 2450 (3)
2. a = 22 and d = 6
   Sn  =  ⁿ/₂ [2a + (n − 1)d] n
   ⁿ/₂ [2 × 22 + (n − 1)6] = 1870
   19n + 3n²  = 1870
   3n²  + 19n − 1870 = 0
   (3n + 85)(n − 22) = 0 3
   ∴ n = 22
   n cannot be a negative because it is the number of terms (2)
3. 1. T_n = –3 + (n − 1)4
      4n – 7 = T_n
   2. T₄  = 5 + 4 = 9; T5  = 9 + 4 = 13; 3 T₆  = 13 + 4 = 17 and
      T₇  = 17 + 4 = 21
   3. 0; 1; 2; 0; 1; 2; 0
   4. T_n = –3 + 12 (n − 1)
      393 = 12n – 15
      12n = 393 + 15 = 408
      n = 34
      S34  = ³⁴/₂ × (–3 + 393)
      = 17 × 390
      = 6630 (10)
4. S₅  = ⁵/₂ ( 1 + 5 )  = 15
   S₄  = ⁴/₂ ( 1+ 4 )  = 10
   T₅ = 15 – 10 = 5 (3)
5. T₂ – T₁ = T₃ – T₂
   2x – (3x + 1) = (3x – 7) – 2x
   2x – 3x – 1 = 3x – 7 – 2x
   –2x + 6 = 0
   2x = 6
   x = 3  (3)
6. 1. T_n = a + (n – 1)d
      T₁₁ = 10 + (11 – 1)(–4)
      = –30
   2. Sn =ⁿ/₂ [2a + (n – 1)d]
      –560 = ⁿ/₂ [2(10) + (n – 1)(–4)]
      –1120 = –4n² + 24n
      4n² – 24n – 1120 = 0
      n² – 6n – 280 = 0
      (n – 20)(n + 14) = 0
      n = 20 or n = –14
      n = 20 only because number of terms cannot be a negative number (6)
      [27] 

1. a = 3 and r = ⁶/₃ =  ¹²/₆  = 2
   Sn =  a( r n  − 1)
   r – 1
   S10 = 3( 2 10  − 1) = 3(1024 − 1) = 3069 (2)
   2 – 1
2. a = 2 and r = ⁶/₂  = ¹⁸/₆ = 3
   Sn =  2(3ⁿ  − 1) = 728
   3 – 1
   2(3ⁿ  − 1)  = 728
   2
   3n − 1 = 728
   3n = 729 = 3⁶
   ∴ n = 6 (3)
   [5] 

1. The question asks you to find the sum of the terms from n = 4 to
   n = 70 if the nth term is 2n – 4.
   a = T₁ = 2(4) – 4 = 4 Find the first term a
   T₂ = 2(5) – 4 = 6
   T₃ = 2(6) – 4 = 8
   So the sequence is 4; 6; 8; … and this is an arithmetic series.
   To check d, calculate T2 – T1
   d = T₂ – T₁ = 6 – 4 = 2
   n = (70 – 4) + 1 = 67 There are 67 terms
   Now we can substitute these values into the formula to find the sum of 67 terms.
   Sn = ⁿ/₂ [2a + (n – 1)d]
   S₆₇ = ⁶⁷/₂ [2(4) + (67 – 1)2]
   S₆₇ = 33.5 [8 + 132] = 4690
   
   (3)
2. This is a geometric series because 5(3)k–1 has the form ar^k–1, T₁ = 5(3)^1–1 = 5 ;
   T₂ = 5(3)²–1 = 15; T₃ = 5(3)^3–1 = 45
   a = 5; r = 3; n = m and Sm  =  65
   S_n = a( rⁿ −1)… substitute
   r – 1
   65 = 5(3^m  − 1) … multiply through by 2
   2
   65 = 5(3^m  − 1)
   2
   130 = 5.3^m – 5 … add like terms
   135 = 5.3^m … divide through by 5
   27 = 3^m … write 27 as a power of 3
   3³ = 3^m … bases are the same, so the powers are equal
   ∴ m = 3 (4)
3. 1. T₁, T₃ and T₅ form a sequence with a common ratio of ½, so T₇ is ¹/₁₆ .
      T₂, T₄ and T₆ form a sequence with a common difference of 3, so T₈ is 13.
   2. S₅₀ = 25 terms of 1st sequence + 25 terms of 2nd sequence
      S₅₀ = ( ½ + ¼ + ¹/₈ + … to 25 terms) + (4 + 7 + 10 + 13 + … to 25 terms)
      

(5)
[12] 

1. T₁ = 8(4)^{1 – 1} = 8 = a
   To find r, find the common ratio using T₁ and T₂, T₂ and T₃.
   T₂ = 8(4)^{1 – 2} = 8(4)–1 = 8 × ¼ = 2
   T₃ = 8(4)^{1 – 3} = 8(4)^–2 = 8 × ¹/₁₆ = ½
   T₂ ÷ T1 = ²/₈ =  ¼  and T3 ÷ T2 =½ = ½ x ½ = ¼
   2
   so r = ¼ and a = 8
   ∴S∞ =   a     =    8    =  8
   1 – r      1 – ¼    ³/₄
   = 8 x ⁴/₃ = ³²/₃
   When dividing by a fraction, you can multiply by the inverse
   ∴S∞ = ³²/₃ or 10²/₃ (3)
2. This is a geometric series with r = 2x – 3
   To converge –1  <  r  <  1
   –1  <  2x – 3  <  1 Add 3 to both sides
   2 < 2x < 4 Divide by 2 on both sides
   1 < x < 2 3 x ≠ ³/₂ (4)
   The series will converge for 1 < x < 2
3. a = 3; r = 2; S_m = 93
   S_n  =  a(1 −  r n )
   1 – r
   93 = 3 (1 − 2^m )
   1 – 2
   93 = 3( 1 − 2^m)
   – 1
   –93 = 3(1 – 2^m)
   –31 = 1 – 2^m
   2^m = 32
   2^m = 2 5
   ∴ m = 5 (4)
4. r = 4x – 1
   –1 < r < 1
   -1 < 4x -1 < 1; x ≠ ¼
   0 < 4x < 2
   0 < x < ½ (3)
   [14]