MATHEMATICS
Grade 11
MORE EXERCISES
Quadratic number patterns, quadratic sequences, answers
  
  
Answers  1
8 ; 13; 20; 29; . . .
     1.1
  

    1.1   The next two terms are 40 and 53 as
            shown in the diagram above.  
                                                                   [ Q 1.1 ]
    
    
$$ \hspace*{2 mm}\mathrm{1.2\kern3mmFirst\ differences\ :\ \ T_{2} − T_{1} = 13 − 8 = 5\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{T_{3} − T_{2} = 20 − 13 = 7\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{T_{4} − T_{3} = 29 − 20 = 9\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Second\ differences\ :\ \ 9 −7 = 7 − 5 = 2\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Second\ differences\ =\ 2a\kern2mm\ } $$
$$ \hspace*{32 mm}\mathrm{2a = 2\ and\ thus\ a = 1\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{First\ first\ difference = 3a + b\kern2mm\ } $$
$$ \hspace*{29 mm}\mathrm{3(1) + b = 5\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{b = 2\kern2mm\ } $$
$$ \hspace*{37 mm}\mathrm{T_{1} = a + b + c\kern2mm\ } $$
$$ \hspace*{23 mm}\mathrm{(1) + (2) + c = 8\kern2mm\ } $$
$$ \hspace*{39mm}\mathrm{c = 5\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{T_{n} = n^2 + 2n + 5\kern2mm\ } $$     [ Q 1.2 ]
    
$$ \hspace*{2 mm}\mathrm{1.3\kern3mmT_{n} = n^2 + 2n + 5\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{T_{21} = (21)^2 + 2(21) + 5\kern2mm\ } $$
$$ \hspace*{15 mm}\mathrm{= 488\kern2mm\ } $$                                  [ Q 1.3 ]
    
$$ \hspace*{2 mm}\mathrm{1.4\kern3mmT_{n} = n^2 + 2n + 5\kern2mm\ } $$
$$ \hspace*{16 mm}\mathrm{n^2 + 2n + 5 = 148\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{n^2 + 2n − 143 = 0\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{(n + 13)(n − 11) = 0\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{n + 13 = 0\ \ \ of\ \ \ n − 11 = 0\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{n = −13\ \ \ of\ \ \ n = 11\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{T_{11} = 148\kern2mm\ } $$                                   [ Q 1.4 ]
-- end #1 -->
  
Answers  2
─2 ; ─1; 4; 13; . . .
     2.1
  

    2.1   The next two terms are 22 and 33 as
             shown in the diagram above.       [ Q 2.1 ]
    
$$ \hspace*{2 mm}\mathrm{2.2\kern3mmFirst\ difference\ :\kern2mm\ } $$
$$ \hspace*{20 mm}\mathrm{T_{2} − T_{1} = (−1) − (−2) = 1\kern2mm\ } $$
$$ \hspace*{20 mm}\mathrm{T_{3} − T_{2} = 4 − (−21) = 5\kern2mm\ } $$
$$ \hspace*{20 mm}\mathrm{T_{4} − T_{3} = 13 − 4 = 9\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Second\ differences\ :\ \ 9 −5 = 5 − 1 = 4\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Second\ differences\ =\ 2a\kern2mm\ } $$
$$ \hspace*{32 mm}\mathrm{2a = 4\ and\ thus\ a = 2\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{First\ first\ difference = 3a + b\kern2mm\ } $$
$$ \hspace*{29 mm}\mathrm{3(2) + b = 1\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{b = −5\kern2mm\ } $$
$$ \hspace*{37 mm}\mathrm{T_{1} = a + b + c\kern2mm\ } $$
$$ \hspace*{21 mm}\mathrm{(2) + (−5) + c = −2\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{c = 1\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{T_{n} = 2n^2 − 5n + 1\kern2mm\ } $$                    [ Q 2.2 ]
    
$$ \hspace*{2 mm}\mathrm{2.3\kern3mmT_{n} = 2n^2 − 5n + 1\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{T_{15} = 2(15)^2 − 2(15) + 1\kern2mm\ } $$
$$ \hspace*{15 mm}\mathrm{= 376\kern2mm\ } $$                                    [ Q 2.3 ]
    
$$ \hspace*{2 mm}\mathrm{2.4\kern3mmT_{n} = 2n^2 − 5n + 1\kern2mm\ } $$
$$ \hspace*{16 mm}\mathrm{2n^2 − 5n + 1 = 494\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{2n^2 − 5n − 493 = 0\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{(2n + 29)(n − 17) = 0\kern2mm\ } $$
$$ \hspace*{15 mm}\mathrm{2n + 29 = 0\ \ \ of\ \ \ n − 17 = 0\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{n = −14,5\ \ \ of\ \ \ n = 17\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{T_{17} = 494\kern2mm\ } $$                      [ Q 2.4 ]
-- end #2 -->
  
Answers  3
patroon : ─5;   5;   17;   31;   . . .
     3.1
  

    3.1   The next two terms are 47 and 71 as
             sown in the diagram above.          [ Q 3.1 ]
    
$$ \hspace*{2 mm}\mathrm{3.2\kern3mmFirst\ differences\ :\ \ T_{2} − T_{1} = 5 − (−5) = 10\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{T_{3} − T_{2} = 17 − (5) = 12\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{T_{4} − T_{3} = 31 − 17 = 14\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Second\ differences\ :\ \ 14 −12 = 12 − 10 = 2\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Second\ differences\ =\ 2a\kern2mm\ } $$
$$ \hspace*{32 mm}\mathrm{2a = 2\ and\ thus\ a = 1\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{First\ first\ difference = 3a + b\kern2mm\ } $$
$$ \hspace*{29 mm}\mathrm{3(1) + b = 10\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{b = 7\kern2mm\ } $$
$$ \hspace*{37 mm}\mathrm{T_{1} = a + b + c\kern2mm\ } $$
$$ \hspace*{23 mm}\mathrm{(1) + (7) + c = −5\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{c = −13\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{T_{n} = n^2 + 7n − 13\kern2mm\ } $$                  [ Q 3.2 ]
    
$$ \hspace*{2 mm}\mathrm{3.3\kern3mmT_{n} = n^2 + 7n − 13\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{T_{9} = (9)^2 + 7(9) − 13\kern2mm\ } $$
$$ \hspace*{13 mm}\mathrm{= 131\kern2mm\ } $$                                    [ Q 3.3 ]
    
$$ \hspace*{2 mm}\mathrm{3.4\kern3mmT_{n} = n^2 + 7n − 13\kern2mm\ } $$
$$ \hspace*{14 mm}\mathrm{n^2 + 7n − 13 = 355\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{n^2 + 7n − 368 = 0\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{(n + 16)(n − 23) = 0\kern2mm\ } $$
$$ \hspace*{15 mm}\mathrm{n − 16 = 0\ \ \ of\ \ \ n + 23 = 0\kern2mm\ } $$
$$ \hspace*{21 mm}\mathrm{n = 16\ \ \ of\ \ \ n = −23\kern2mm\ } $$
$$ \hspace*{18 mm}\mathrm{T_{16} = 355\kern2mm\ } $$                          [ Q 3.4 ]
  
Answers  4
12;   7;   ─2;   ─15;   . . .
     4.1
  

    4.1   The next two terms are 22 and 33 as
             shown in the diagram above.          [ Q 4.1 ]
    
$$ \hspace*{2 mm}\mathrm{4.2\kern3mmFirst\ differences\ :\kern2mm\ } $$
$$ \hspace*{20 mm}\mathrm{T_{2} − T_{1} = 7 − 12 = −5\kern2mm\ } $$
$$ \hspace*{20 mm}\mathrm{T_{3} − T_{2} = (−2) − (7) = −9\kern2mm\ } $$
$$ \hspace*{20 mm}\mathrm{T_{4} − T_{3} = (−15) − (−2) = −13\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Second\ differences\ =\ \ −13 −(−9) = (−9) − (−5) = −4\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Second\ differences\ =\ 2a\kern2mm\ } $$
$$ \hspace*{32 mm}\mathrm{2a = −4\ and\ thus\ a = −2\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{First\ first\ difference = 3a + b\kern2mm\ } $$
$$ \hspace*{27 mm}\mathrm{3(−2) + b = −5\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{b = 1\kern2mm\ } $$
$$ \hspace*{37 mm}\mathrm{T_{1} = a + b + c\kern2mm\ } $$
$$ \hspace*{23 mm}\mathrm{(−2) + (1) + c = 12\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{c = 13\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{T_{n} = −2n^2 + n + 13\kern2mm\ } $$                  [ Q 4.2 ]
    
$$ \hspace*{2 mm}\mathrm{4.3\kern3mmT_{n} = −2n^2 + n + 13\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{T_{11} = −2(11)^2 + (11) + 13\kern2mm\ } $$
$$ \hspace*{13 mm}\mathrm{= −218\kern2mm\ } $$                                    [ Q 4.3 ]
    
$$ \hspace*{2 mm}\mathrm{4.4\kern3mmT_{n} = −2n^2 + n + 13\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{−2n^2 + n + 13 = −617\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{−2n^2 + n + 630 = 0\kern2mm\ } $$
$$ \hspace*{14 mm}\mathrm{2n^2 − n − 630 = 0\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{(2n + 35)(n − 18) = 0\kern2mm\ } $$
$$ \hspace*{15 mm}\mathrm{2n + 35 = 0\ \ \ or \ \ n − 18 = 0\kern2mm\ } $$
$$ \hspace*{21 mm}\mathrm{n = −17,5\ \ \ or\ \ \ n = 18\kern2mm\ } $$
$$ \hspace*{18 mm}\mathrm{T_{18} = −617\kern2mm\ } $$                          [ Q 4.4 ]
  
Answers  5
    
$$ \hspace*{2 mm}\mathrm{5.\kern3mmFirst\ second\ difference\ \ :\ \ 8 − y = 6\kern2mm\ } $$
$$ \hspace*{54 mm}\mathrm{y = 2\kern2mm\ } $$
     Let p = first first difference.
$$ \hspace*{24 mm}\mathrm{6 − p = 2\kern2mm\ } $$
$$ \hspace*{29 mm}\mathrm{p = 4\kern2mm\ } $$
$$ \hspace*{24 mm}\mathrm{x + 4 = y\kern2mm\ } $$
$$ \hspace*{29 mm}\mathrm{x = 2 − 4\kern2mm\ } $$
$$ \hspace*{29 mm}\mathrm{x = −2\kern2mm\ } $$
$$ \hspace*{19 mm}\mathrm{x = −2\ \ and\ \ y = 2\kern2mm\ } $$           [ V 5. ]
$$ \hspace*{30 mm}\mathrm{\bold{OR}\kern2mm\ } $$
     5.
  

    5.   x = −2 and y = 2 as shown in the diagram
             diagram above.          [ Q 5. ]
  
Answers  6
    
      1;   p;   21;   q; . . .
$$ \hspace*{2 mm}\mathrm{6.\kern3mmT_{2} − T_{1} = p − 1\kern2mm\ } $$
$$ \hspace*{8 mm}\mathrm{T_{3} − T_{2} = 21 − p\kern2mm\ } $$
$$ \hspace*{8 mm}\mathrm{First\ second\ difference\ :\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{(21 − p) − (p − 1) = 4\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{(22 − 2p) = 4\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{2p = 18\kern2mm\ } $$
$$ \hspace*{14 mm}\mathrm{p = 9\kern2mm\ } $$
$$ \hspace*{8 mm}\mathrm{Second\ first\ difference\ = 21 − 9\kern2mm\ } $$
$$ \hspace*{44 mm}\mathrm{= 12\kern2mm\ } $$
$$ \hspace*{11 mm}\mathrm{Third\ first\ difference\ = 12 + 4\kern2mm\ } $$
$$ \hspace*{44 mm}\mathrm{= 16\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{q − 21 = 16\kern2mm\ } $$
$$ \hspace*{40 mm}\mathrm{q = 37\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{p = 9\ \ and\ \ q = 37\kern2mm\ } $$                 [ Q 6. ]
  
Answers  7
    
─ 2;   x;   y;   . . .
$$ \hspace*{2 mm}\mathrm{7.\kern3mmFirst\ first\ difference\ ;\ 7 − 2 = 5\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{ x − (−2) = 5\kern2mm\ } $$
$$ \hspace*{21 mm}\mathrm{x = 3\kern2mm\ } $$
$$ \hspace*{15 mm}\mathrm{y − x = 7\kern2mm\ } $$
$$ \hspace*{20 mm}\mathrm{y = 10\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{x = 3\ \ and\ \ y = 10\kern2mm\ } $$                 [ Q 7. ]
$$ \hspace*{30 mm}\mathrm{\bold{OR}\kern2mm\ } $$
     7.
  

    7.   x = 3 and y = 10 as shown in the diagram
             diagram above.          [ Q 7. ]
  
Answers  8
    
          4;    9;    x;    37;   . . .
$$ \hspace*{2 mm}\mathrm{8.1\kern3mmT_{2} − T_{1} = 9 − 4 = 5\kern2mm\ } $$
$$ \hspace*{8 mm}\mathrm{T_{3} − T_{2} = x − 9\kern2mm\ } $$
$$ \hspace*{8 mm}\mathrm{T_{4} − T_{3} = 37 − x\kern2mm\ } $$
$$ \hspace*{8 mm}\mathrm{Calculate\ second\ differences\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{(x − 9) − 5 = (37 − x) − (x − 9)\kern2mm\ } $$
$$ \hspace*{18 mm}\mathrm{x − 14 = 46 − 2x\kern2mm\ } $$
$$ \hspace*{23 mm}\mathrm{3x = 60\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{x = 20\kern2mm\ } $$                    [ Q 8.1 ]
     8.2   Patroon is nou :  4;    9;    20;    37;   . . .
$$ \hspace*{10 mm}\mathrm{T_{2} − T_{1} = 9 − 4 = 5\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{3} − T_{2} = 20 − 9 = 11\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{Second\ difference\ =\ 2a\kern2mm\ } $$
$$ \hspace*{30 mm}\mathrm{2a = 11 − 5 = 6\kern2mm\ } $$
$$ \hspace*{32 mm}\mathrm{a = 3\kern2mm\ } $$
$$ \hspace*{8 mm}\mathrm{First\ first\ difference\ = 3a + b\kern2mm\ } $$
$$ \hspace*{29 mm}\mathrm{3(3) + b = 5\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{b = −4\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{First\ term\ = a + b + c\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{(3) + (−4) + c = 4\kern2mm\ } $$
$$ \hspace*{28 mm}\mathrm{c = 5\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{T_{n} = 3n^2 − 4n + 5\kern2mm\ } $$
                                                                   [ Q 8.2 ]
  
Answers  9
     9.1   Patroon is :  x;    0;    6;    y;    24  . . .
$$ \hspace*{10 mm}\mathrm{T_{3} − T_{2} = 6 − 0 = 6\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{4} − T_{3} = y − 6\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{5} − T_{4} = 24 − y\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{Constant\ second\ difference\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{(24 − y) −(y − 6) = (y − 6) − (6 − 0)\kern2mm\ } $$
$$ \hspace*{23 mm}\mathrm{30 − 2y = y − 12\kern2mm\ } $$
$$ \hspace*{31 mm}\mathrm{3y = 42\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{y = 14\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{(24 − y) −(y − 6) = (y − 6) − (6 − 0)\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{(24 − 14) −(14 − 6) = (14 − 6) − (6 − 0)\kern2mm\ } $$
$$ \hspace*{29 mm}\mathrm{10 −8 = 8 − 6 = 2\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{Second\ difference\ = 2\kern2mm\ } $$                [ Q 9.1 ]
     9.2 $$ \hspace*{10 mm}\mathrm{Second\ differences\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{(6 − 0) −(0 − x) = 2\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{6 + x = 2\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{x = −4\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{First\ term\ = −4\kern2mm\ } $$                [ Q 9.2 ]

  
Answers  10
0      10.1     6;    7;    12;    p;    . . .
  
$$ \hspace*{10 mm}\mathrm{T_{2} − T_{1} = 7 − 6 = 1\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{3} − T_{2} = 12 − 7 = 5\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{4} − T_{3} = p − 12\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{Constant\ second\ differences\ = 5 − 1 = 4\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{(p − 12) −(12 − 7) = 4\kern2mm\ } $$
$$ \hspace*{43 mm}\mathrm{p − 17 = 4\kern2mm\ } $$
$$ \hspace*{50 mm}\mathrm{p = 21\kern2mm\ } $$
                                                                 [ Q 10.1 ]
    
    6;    7;    12;    p;    . . .
  
$$ \hspace*{2 mm}\mathrm{10.2\kern3mmSecond differences\ :\ 2a = 4\kern2mm\ } $$
$$ \hspace*{40 mm}\mathrm{a = 2\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{First\ first\ difference\ = 3a + b\kern2mm\ } $$
$$ \hspace*{31 mm}\mathrm{3(2) + b = 1\kern2mm\ } $$
$$ \hspace*{41 mm}\mathrm{b = −5\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{First\ term = a + b + c\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{(2) + (−5) + c = 6\kern2mm\ } $$
$$ \hspace*{28 mm}\mathrm{c = 9\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{n} = 2n^2 − 5n + 9\kern2mm\ } $$                  [ Q 10.2 ]
    
$$ \hspace*{2 mm}\mathrm{10.3\kern3mmT_{n} = 2n^2 − 5n + 9\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{Consecutive\ terms\ differ\ by\ 53\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{\therefore T_{n} − T_{n − 1} = 53\kern2mm\ } $$
$$ \hspace*{7 mm}\mathrm{2n^2 − 5n + 9 − {2(n − 1)^2 − 5(n − 1) + 9} = 53\kern2mm\ } $$
$$ \hspace*{2 mm}\mathrm{2n^2 − 5n + 9 − {2(n^2 − 2n + 1) − 5n + 5 + 9} = 53\kern2mm\ } $$
$$ \hspace*{2 mm}\mathrm{2n^2 − 5n + 9 − 2n^2 + 4n − 2 + 5n − 5 − 9 = 53\kern2mm\ } $$
$$ \hspace*{52 mm}\mathrm{4n − 7 = 53\kern2mm\ } $$
$$ \hspace*{60 mm}\mathrm{n = 15\kern2mm\ } $$
$$ \hspace*{2 mm}\mathrm{The\ terms\ are\ \ T_{14}\ and\ T_{15}\kern2mm\ } $$
$$ \hspace*{2 mm}\mathrm{T_{14} = 2(14)^2 −(5 \times; 14) + 9 = 331\kern2mm\ } $$
$$ \hspace*{2 mm}\mathrm{T_{15} = 2(15)^2 −(5 \times; 15) + 9 = 384\kern2mm\ } $$
$$ \hspace*{2 mm}\mathrm{T_{15} − T_{14} = 384 − 331 = 53\kern2mm\ } $$              [ Q 10.3 ]

  
Answers  11
     $$ \hspace*{2 mm}\mathrm{11.\kern3mmT_{n} = 3(n + 2)^2 − 4\kern2mm\ } $$
$$ \hspace*{14mm}\mathrm{= 3n^2 + 12n + 12 − 4\kern2mm\ } $$
$$ \hspace*{14 mm}\mathrm{= 3n^2 + 12n + 8\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{\therefore a = 3\ \ b = 12\ \ and\ \ c = 8\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{First\ first\ difference = 3a + b\kern2mm\ } $$
$$ \hspace*{43 mm}\mathrm{= 3(3) + (12)\kern2mm\ } $$
$$ \hspace*{43 mm}\mathrm{= 21\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{Constant\ second\ difference = 2a\kern2mm\ } $$
$$ \hspace*{50 mm}\mathrm{= 3(2)\kern2mm\ } $$
$$ \hspace*{50 mm}\mathrm{= 6\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{First\ term\ = a + b + c\kern2mm\ } $$
$$ \hspace*{30 mm}\mathrm{= (2) + (12) + (8)\kern2mm\ } $$
$$ \hspace*{30 mm}\mathrm{= 23\kern2mm\ } $$                    [ Q 11. ]
$$ \hspace*{40 mm}\mathrm{\bold{OR}\kern2mm\ } $$ $$ \hspace*{10 mm}\mathrm{T_{n} = 3(n + 2)^2 − 4\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{1} = 3(1 + 2)^2 − 4 = 23\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{2} = 3(2 + 2)^2 − 4 = 44\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{3} = 3(3 + 2)^2 − 4 = 71\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{First\ first\ difference\ = T_{2} − T_{1}\kern2mm\ } $$
$$ \hspace*{44 mm}\mathrm{= 44 − 23\kern2mm\ } $$
$$ \hspace*{44 mm}\mathrm{= 21\kern2mm\ } $$
$$ \hspace*{2 mm}\mathrm{Constant\ second\ difference = (T_{3} − T_{2}) − (T_{2} − T_{1})\kern2mm\ } $$
$$ \hspace*{42 mm}\mathrm{= (71 − 44) − (44 − 23)\kern2mm\ } $$
$$ \hspace*{42 mm}\mathrm{= 27 − 21\kern2mm\ } $$
$$ \hspace*{42 mm}\mathrm{= 6\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{First\ term\ = 23\kern2mm\ } $$                   [ Q 11. ]

  
Answers  12
    
$$ \hspace*{2 mm}\mathrm{12.1\kern3mmFirst\ differences\ :\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{2} − T_{1} = −8 − (−7) = −1\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{3} − T_{2} = −11 − (−8) = −3\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{4} − T_{3} = −16 − (−11) = −5\kern2mm\ } $$
$$ \hspace*{8 mm}\mathrm{Constant\ second\ differences\ = −3 − (−1)\kern2mm\ } $$
$$ \hspace*{54 mm}\mathrm{= −2\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{2a = −2\ and\ thus\ a = −1\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{First\ first\ difference = 3a + b\kern2mm\ } $$
$$ \hspace*{28 mm}\mathrm{3(−1) + b = −1\kern2mm\ } $$
$$ \hspace*{40 mm}\mathrm{b = 2\kern2mm\ } $$
$$ \hspace*{27 mm}\mathrm{First\ term = a + b + c\kern2mm\ } $$
$$ \hspace*{24 mm}\mathrm{(−1) + 2 + c = −7\kern2mm\ } $$
$$ \hspace*{40 mm}\mathrm{c = −8\kern2mm\ } $$
$$ \hspace*{20 mm}\mathrm{T_{n} = −n^2 + 2n − 8\kern2mm\ } $$
                                                                 [ Q 12.1 ]
    
$$ \hspace*{2 mm}\mathrm{12.2\kern3mmT_{n} = −n^2 + 2n − 8\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{17} = −17^2 + 2(17) − 8\kern2mm\ } $$
$$ \hspace*{16 mm}\mathrm{= −263\kern2mm\ } $$
                                                                 [ Q 12.2 ]
    
$$ \hspace*{2 mm}\mathrm{12.3\kern3mmT_{n} = −n^2 + 2n − 8\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{−n^2 + 2n − 8 < −313\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{n^2 − 2n − 305 > 0\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{n = \frac{−(−2) \plusmn \sqrt((−2)^2 − 4(−1)(305))}{2(−1)}\kern2mm\ } $$

$$ \hspace*{13 mm}\mathrm{= \frac{2 \plusmn \sqrt(4 + 1220)}{−2}\kern2mm\ } $$

$$ \hspace*{13 mm}\mathrm{n = 18,4929\ \ OR\ \ n = −16,4949\kern2mm\ } $$
$$ \hspace*{13 mm}\mathrm{n = 18, \kern3mm\ T_{18} = −276\kern2mm\ } $$
                                                                 [ Q 12.3 ]
    
$$ \hspace*{2 mm}\mathrm{12.4\kern3mmTn − T(n − 1) = −63\kern2mm\ } $$
$$ \hspace*{6 mm}\mathrm{−n^2 + 2n − 8 − \kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{(−(n − 1)^2 + 2(n − 1) − 8) = −63\kern2mm\ } $$
$$ \hspace*{6 mm}\mathrm{−n^2 + 2n − 8 − \kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{(−(n^2 − 2n + 1) + 2n − 2 − 8) = −63\kern2mm\ } $$
$$ \hspace*{6 mm}\mathrm{−n^2 + 2n − 8 − \kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{(−n^2 + 2n − 1 + 2n − 2 − 8) = −63\kern2mm\ } $$
$$ \hspace*{6 mm}\mathrm{−n^2 + 2n − 8 + \kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{n^2 − 2n + 1 − 2n + 2 + 8) = −63\kern2mm\ } $$
$$ \hspace*{6 mm}\mathrm{−2n + 3 = −63\kern2mm\ } $$
$$ \hspace*{6 mm}\mathrm{−2n = −66\kern2mm\ } $$
$$ \hspace*{6 mm}\mathrm{n = 33\kern2mm\ } $$
      The two terms are T32 and T33,
      i.e. −968 and −1 031                      [ Q 12.4 ]
$$ \hspace*{36 mm}\mathrm{\bold{OR}\kern2mm\ } $$
    12.4  The difference between Tn and Tn − 1
              is given by the nth term of the linear
              sequence representing the first
              differences.
              Create the pattern and solve for
              Tn = −63.
              Quadratic pattern :   −7   −8   −11   −16
              First difference :           −1   −3     −5
$$ \hspace*{6 mm}\mathrm{First\ term = a = −1\ and\kern2mm\ } $$ $$ \hspace*{6 mm}\mathrm{common\ difference\ d = −2\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{n} = a + (n − 1)d\kern2mm\ } $$ $$ \hspace*{15 mm}\mathrm{= −1 + (n − 1)(−2)\kern2mm\ } $$
$$ \hspace*{15 mm}\mathrm{= −2n + 1\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{∴ −2n + 1 = −63\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{n = 32\kern2mm\ } $$
      The two terms are T32 and T33,
      i.e. −968 and −1 031                      [ Q 12.4 ]
  
Answers  13
    
$$ \hspace*{2 mm}\mathrm{13.1\kern3mmT_{n} = −n^2 + bn − 150\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{First\ first\ difference = 15\ and\ a = −1 \kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{∴ 3a + b = 15\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{3(−1) + b = 15\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{b = 18\kern2mm\ } $$                                      [ Q 13.1 ]
    
$$ \hspace*{2 mm}\mathrm{13.2\kern3mmT_{n} = −n^2 + 18n − 150\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{16} = −16^2 + 18 \times 16 − 150\kern2mm\ } $$
$$ \hspace*{16 mm}\mathrm{= −118\kern2mm\ } $$                             [ Q 13.2 ]
    
$$ \hspace*{2 mm}\mathrm{13.3\kern3mmT_{n} = −n^2 + 18n − 150\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{−n^2 + 18n − 150 = −598\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{n^2 − 18n − 448 = 0\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{n = \frac{−(−18) \plusmn (\sqrt{(−18)^2 − 4 \times 1 \times (−448)})}{2 \times 1}\kern2mm\ } $$

$$ \hspace*{12 mm}\mathrm{= \frac{18 \plusmn \sqrt{324 + 1792}}{2}\kern2mm\ } $$

$$ \hspace*{12 mm}\mathrm{= \frac{18 \plusmn 46}{2}\kern2mm\ } $$

$$ \hspace*{12 mm}\mathrm{n = 32\ \ or\ \ n = −14\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{T_{32} = −598\kern2mm\ } $$                             [ Q 13.3 ]
    
$$ \hspace*{2 mm}\mathrm{13.4\kern3mmT_{n} = −n^2 + 18n − 150\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{−n^2 + 18n − 150 − (−(n − 1)^2 + 18(n − 1) − 150 ) 270\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{n^2 − 18n − 120 > 0\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{x = \frac{−(−18) \plusmn (\sqrt{(−18)^2 − 4 \times 1 \times (−120)})}{2 \times 1}\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{x = 23,1774\ \ or\ \ x = −5,1774\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{n > 23,1774\ \ or\ \ n < −5,1774\kern2mm\ } $$
$$ \hspace*{8 mm}\mathrm{\therefore; n =24 and\ T_{24} < −270,\ \kern3mm\ T_{24} = −294\kern2mm\ } $$
                                                               [ Q 13.4 ]
    
        13.5Calculate the value of the first
              four terms and the first differences
$$ \hspace*{10 mm}\mathrm{T_{n} = −n^2 + 18n − 150\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{1} = −(1)^2 + 18(1) − 150 = − 133\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{2} = −(2)^2 + 18(2) − 150 = − 118\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{3} = −(3)^2 + 18(3) − 150 = − 105\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{4} = −(4)^2 + 18(4) − 150 = − 94\kern2mm\ } $$
$$ \hspace*{6 mm}\mathrm{First\ first\ difference = T_{2} − T_{1}\kern2mm\ } $$
$$ \hspace*{37 mm}\mathrm{= −118 − (−133) = 15\kern2mm\ } $$
$$ \hspace*{6 mm}\mathrm{Second\ first\ difference = T_{3} − T_{2}\kern2mm\ } $$
$$ \hspace*{42 mm}\mathrm{= −105 − (−118) = 13\kern2mm\ } $$
$$ \hspace*{6 mm}\mathrm{Third\ first\ difference = T_{4} − T_{3}\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= −94 − (−105) = 11\kern2mm\ } $$
$$ \hspace*{20 mm}\mathrm{\bold{OR}\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{First\ difference\ (FD) = T_{n + 1} − T_{n}\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{First\ FD = T_{2} − T_{1}\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{= {−(2)^2 + 18(2) − 150} − (−(1)^2 + 18() − 150)\kern2mm\ } $$
$$ \hspace*{22 mm}\mathrm{= (−118) − (−133) = 15\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{Second\ FD = T_{3} − T_{3}\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{= {−(3)^2 + 18(3) − 150} − (−(2)^2 + 18(2) − 150)\kern2mm\ } $$
$$ \hspace*{22 mm}\mathrm{= (−105) − (−118) = 13\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{Third\ FD = T_{4} − T_{3}\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{= {−(4)^2 + 18(4) − 150} − (−(3)^2 + 18(3) − 150)\kern2mm\ } $$
$$ \hspace*{22 mm}\mathrm{= (−94) − (−105) = 11\kern2mm\ } $$
              Now write down the general term
              of the pattern of the first differences :
              a = 15 and d = 13 - 15 = 11 - 13 = -2
$$ \hspace*{11 mm}\mathrm{T_{n} = a + (n − 1)d = 15 + (n − 1)(−2)\kern2mm\ } $$
$$ \hspace*{16 mm}\mathrm{= −2n + 17\kern2mm\ } $$
                                                               [ Q 13.5 ]
$$ \hspace*{40 mm}\mathrm{\bold{OR}\kern2mm\ } $$
$$ \hspace*{2 mm}\mathrm{13.5\kern3mmFirst\ difference = T_{n} −T_{n − 1}\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{= −n^2 + 18n − 150 − (−(n − 1)^2 + 18(n − 1) − 150 )\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{= −2n + 17\ \ . . . \ see 12.4\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{n} = −2n + 17\kern2mm\ } $$
$$ \hspace*{20 mm}\mathrm{\bold{OR}\kern2mm\ } $$
$$ \hspace*{2 mm}\mathrm{13.5\kern3mmFirst\ difference = T_{n + 1} −T_{n}\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{= −(n + 1)^2 + 18(n + 1) − 150 − (−n^2 + 18n − 150 )\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{= −2n + 17\ \ . . . \ see 12.4\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{n} = −2n + 17\kern2mm\ } $$
                                                               [ Q 13.5 ]
    
$$ \hspace*{2 mm}\mathrm{13.6\kern3mmT_{n + 1} −T_{n} = −5\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{−2n + 17 = −5\kern2mm\ } $$
$$ \hspace*{23 mm}\mathrm{n = 11\kern2mm\ } $$
              The terms are T12 and T11..
$$ \hspace*{10 mm}\mathrm{T_{12} = −(12)^2 + 18(12) − 150 = − 78\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{11} = −(11)^2 + 18(11) − 150 = − 73\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{T_{12} − T_{11} = − 78 − (− 73)\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{ = − 5\kern2mm\ } $$                      [ Q 13.6 ]
  
Answers  14
    
$$ \hspace*{2 mm}\mathrm{14.1\kern3mmT_{n} = 3n^2 − 4n + c\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{First\ term = −14\ \ a = 3\ and\ b = −4 \kern2mm\ } $$
$$ \hspace*{24 mm}\mathrm{∴ a + b + c = −14\kern2mm\ } $$
$$ \hspace*{23 mm}\mathrm{3 + (−4) + c = −14\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{c = −13\kern2mm\ } $$
                                                                    [ Q 14.1 ]
    
$$ \hspace*{2 mm}\mathrm{14.2\kern3mmT_{n} = 3n^2 − 4n − 13\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{T_{7} = 3(7)^2 − 4(7) − 13\kern2mm\ } $$
$$ \hspace*{16 mm}\mathrm{= 106\kern2mm\ } $$                             [ Q 14.2 ]
    
$$ \hspace*{2 mm}\mathrm{14.3\kern3mmT_{n} = 3n^2 − 4n − 13 > 300\kern2mm\ } $$
$$ \hspace*{19 mm}\mathrm{3n^2 − 4n − 13 > 300\kern2mm\ } $$
$$ \hspace*{17 mm}\mathrm{3n^2 − 4n − 313 > 0\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{x = \frac{−(−4) \plusmn \sqrt{(−4)^2 − 4(3)(−313)}}{2(3)}\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{x = \frac{4 \plusmn \sqrt{16 + 3 756}}{6}\kern2mm\ } $$

$$ \hspace*{12 mm}\mathrm{x = \frac{4 \plusmn \sqrt{3 772}}{6}\kern2mm\ } $$

$$ \hspace*{12 mm}\mathrm{x = \frac{4 \plusmn 61,4156}{6}\kern2mm\ } $$

$$ \hspace*{12 mm}\mathrm{x = 10,9028 \ or\ x = -9,5694\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{n = 11\ and\ T_{11} = 306\kern2mm\ } $$         [ Q 14.3 ]
    
$$ \hspace*{2 mm}\mathrm{14.4\kern3mmFirst\ difference (FD) = T_{n + 1} − T_{n}\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{First\ FD = T_{2} − T_{1}\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{= −9 − (−14) = 5\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{Second\ FD = T_{3} − T_{2}\kern2mm\ } $$
$$ \hspace*{31 mm}\mathrm{= 2 − (−9) = 11\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{Third\ FD = T_{4} − T_{3}\kern2mm\ } $$
$$ \hspace*{27 mm}\mathrm{= 19 − (2) = 17\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{T_{n} = a + (n − 1)d\ \ a = 5\ and\kern2mm\ } $$
$$ \hspace*{14 mm}\mathrm{d = 17 − 11 = 11 − 5 = 6\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{T_{n} = 5 + (n − 1) \times 6\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{T_{n} = 6n − 1\kern2mm\ } $$
$$ \hspace*{20 mm}\mathrm{\bold{OR}\kern2mm\ } $$
$$ \hspace*{2 mm}\mathrm{First\ difference (FD) = T_{n + 1} − T_{n}\kern2mm\ } $$
$$ \hspace*{2 mm}\mathrm{= 3(n+1)^2−4(n+1)−13 −(3n^2−4n−13) \kern2mm\ } $$
$$ \hspace*{2 mm}\mathrm{= 6n − 1\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{Calculate\ the\ terms\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{T_{n} = 125\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{6n − 1 = 125\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{n = 21\kern2mm\ } $$
              The terms are T21 and T22..
$$ \hspace*{12 mm}\mathrm{T_{22} − T_{21} = 1 351 − 1 226\kern2mm\ } $$
$$ \hspace*{27 mm}\mathrm{= 125\kern2mm\ } $$                     [ Q 14.3 ]