{"id":2164,"date":"2023-10-02T10:28:38","date_gmt":"2023-10-02T10:28:38","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=2164"},"modified":"2023-10-02T10:42:37","modified_gmt":"2023-10-02T10:42:37","slug":"week-1-ss1-second-term-further-mathematicsnotes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-1-ss1-second-term-further-mathematicsnotes\/","title":{"rendered":"Week 1 – SS1 Second Term Further Mathematics Notes"},"content":{"rendered":"

\u00a0\u00a0\u00a0\u00a0
\n\tSECOND TERM E-LEARNING NOTE <\/strong><\/p>\n

\u00a0<\/p>\n

SUBJECT: FURTHER MATHEMATICS \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 CLASS: SS1 SCHEME OF WORK
\n<\/h3>\n

\u00a0<\/p>\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
WEEK <\/strong><\/td>\nTOPIC <\/strong><\/td>\n<\/tr>\n
1 <\/td>\nArithmetic Progression (AP) <\/td>\n<\/tr>\n
2 <\/td>\nGeometric Progression (GP) <\/td>\n<\/tr>\n
3 <\/td>\nLinear inequalities in one variable <\/td>\n<\/tr>\n
4 <\/td>\nInequalities in two variables (Graph of inequalities) <\/td>\n<\/tr>\n
5 <\/td>\nIntroduction to the concept of functions. <\/td>\n<\/tr>\n
6 <\/td>\nReview of half term work. <\/td>\n<\/tr>\n
7 <\/td>\nFunctions (one \u2013 to \u2013 one, onto, composite and inverse functions) <\/td>\n<\/tr>\n
8 <\/td>\nTrigonometric ratio: Graph of Sine, Cosine and tangent of angles, deviation of trigonometric ratio of special angles (300<\/sup>, 450<\/sup> and 600<\/sup>). Application of trigonometric ratios. <\/td>\n<\/tr>\n
9 <\/td>\nLogical reasoning: Simple True and False statement, Negation, Converse and Contra positive of statement, <\/td>\n<\/tr>\n
10 <\/td>\nLogical reasoning continues: Compound statement, connectives and their symbols, conditional statements and symbols. <\/td>\n<\/tr>\n
11 <\/td>\nRevision of Second Term’s lesson <\/td>\n<\/tr>\n
12 <\/td>\nExamination <\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

\u00a0<\/p>\n

REFERECES
\n<\/h3>\n
    \n
  • FutherMaths Project 1 and 2 by TuttuhAdegun (main text). \u27a2 Additional Mathematics by Godman\n<\/li>\n
  • Further Mathematics by E. Egbe et al.\n<\/li>\n<\/ul>\n

    \u00a0
    \n\u00a0<\/p>\n

    WEEK ONE \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 DATE\u2026\u2026\u2026\u2026\u2026 TOPIC: SEQUENCE & SERIES CONTENT
    \n<\/h3>\n

    \u00a0<\/p>\n

      \n
    • Sequence and series\n<\/li>\n
    • Arithmetic Progression (AP)\n<\/li>\n
    • Arithmetic Mean\n<\/li>\n
    • Sum of terms in an AP\n<\/li>\n<\/ul>\n

      \u00a0<\/p>\n

      Sequence & Series
      \n<\/h4>\n

      A sequence is a pattern of numbers arranged in a particular order. Each of the number in the sequence is called a term. The terms are related to one another according to a well defined rule. <\/p>\n

      \u00a0Consider the sequence 1, 4, 7, 10, 13 \u2026., 1 is the first term,(T1<\/sub>) 4 is the second term(T2<\/sub>), 7 is the third term (T3<\/sub>). The sum of the terms in a sequence is regarded as series. The series of the above sequence is 1 + 4 + 7 + 10 + 13 = 35 <\/p>\n

      \u00a0
      \n\u00a0<\/p>\n

      The nth term of a Sequence
      \n<\/h4>\n

      The nth term of a sequence whose rule is stated may be represented by Tn<\/sub>so that T1<\/sub>, T2<\/sub>, T3<\/sub>etc represent the first term, second term, third term \u2026 etc respectively. Consider the sequence 5, 9, 13, 17, 21 \u2026\u2026..
      \nT1<\/sub> = 5 + 4(0)
      \nT2<\/sub> = 5 + 4(1)
      \nT3<\/sub> = 5 + 4 (2) <\/p>\n

      \u00a0T4 <\/sub>= 5 + 4 (3)
      \nTn<\/sub> = 5 + 4 (n \u2013 1)
      \nTn<\/sub> = 5 + 4n \u2013 4 = 4n +1
      \nwhen n = 30
      \nT30<\/sub> = 4(30) + 1 T30<\/sub> = 121 <\/p>\n

      \u00a0
      \n\u00a0Find the nth term of these sequences: <\/p>\n

      \u00a0<\/p>\n

        \n
      1. 3, 5, 7, 9 \u2026\u2026 2n + 1\n<\/li>\n
      2. 0, 1, 4, 9 \u2026\u2026\u2026 (n -1)2 <\/sup>\n\t\t<\/li>\n
      3. 1\/3, 3\/4, 1, 7\/6 \u2026\u2026\u2026\u2026\u2026\u20262n – 1
        \n\t\t\t<\/sup> \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0n + 2\n<\/li>\n<\/ol>\n

        Examples<\/p>\n<\/h4>\n

        Write down the first four terms of the sequence whose general term is given by:
        \n(i) Tn<\/sub> = n+1 (ii) Tn<\/sub> = 5 x (1<\/sup>\/2<\/sub>)n-2<\/sup>
        \n\t3n +2
        \n \u00a0\u00a0\u00a0\u00a0<\/strong> \u00a0\u00a0\u00a0\u00a0T4<\/sub> = 4 + 1 = 5<\/sup>\/14<\/sub>
        \n\tSolution \u00a0\u00a0\u00a0\u00a0<\/strong> i. \u00a0\u00a0\u00a0\u00a0Tn<\/sub> = n+1 \u00a0\u00a0\u00a0\u00a0
        \n \u00a0\u00a0\u00a0\u00a0 3n + 2 \u00a0\u00a0\u00a0\u00a0(ii) Tn<\/sub> = 5 x (1\/2)n-2<\/sup>
        \n\t \u00a0\u00a0\u00a0\u00a0T1 <\/sub>= 1 + 1 = 2<\/sup>\/5<\/sub> \u00a0\u00a0\u00a0\u00a0T1<\/sub> = 5 x (1\/2)1-2 <\/sup> = 5(1<\/sup>\/2<\/sub>)-1 <\/sup>= 5(2-1<\/sup>)-1<\/sup> = 5 x 2 = 10
        \n \u00a0\u00a0\u00a0\u00a0 3(1) + 2 \u00a0\u00a0\u00a0\u00a0<\/sub>T2<\/sub> = 5 x (1\/2)2-2 <\/sup> = 5(1<\/sup>\/2<\/sub>)0<\/sup> = 5 x 1 = 5
        \n \u00a0\u00a0\u00a0\u00a0T2 <\/sub> = 2 + 1 = 3<\/sup>\/8<\/sub> \u00a0\u00a0\u00a0\u00a0T3<\/sub> = 5 x (1\/2)3-2<\/sup> = 5 x (1<\/sup>\/2<\/sub>) = 5<\/sup>\/2 <\/sub>
        \n\t \u00a0\u00a0\u00a0\u00a0 3(2) +2 \u00a0\u00a0\u00a0\u00a0T4<\/sub> = 5 x (1\/2)4-2<\/sup> = 5(1<\/sup>\/2<\/sub>)2<\/sup> = 5<\/sup>\/4 <\/sub>
        \n\t \u00a0\u00a0\u00a0\u00a0T3<\/sub> = 3 + 1 = 4<\/sup>\/11<\/sub> \u00a0\u00a0\u00a0\u00a0The sequence is 10, 5, 5<\/sup>\/2<\/sub>, 5<\/sup>\/4<\/sub> \u2026\u2026\u2026
        \n \u00a0\u00a0\u00a0\u00a0 3(3) + 2 \u00a0\u00a0\u00a0\u00a0
        \n 3(4) + 2
        \nThe sequence is 2<\/sup>\/5<\/sub>, 3<\/sup>\/8<\/sub>, 4<\/sup>\/11<\/sub>, 5<\/sup>\/14<\/sub> \u2026\u2026.. <\/p>\n

        \u00a0<\/p>\n

        Evaluation
        \n<\/h4>\n

        Find the first term of the sequence whose general term is given by
        \n(i) 50 \u2013 (\u00bd)n<\/sup> (ii) 2 + 3<\/sup>\/2<\/sub>(n+1)<\/sup><\/p>\n

        \u00a0<\/p>\n

        Arithmetic Progression (A.P) or Linear Sequence
        \n<\/h4>\n

        An arithmetic progression (A.P) is generated by adding or subtracting a constant number to a preceding term to get a term. This constant number is called the common difference designated by the letter d. The first term is designated by a. <\/p>\n

        \n\n\n\n
        \n\u00a0
        \n\u00a0
        \n\u00a0T1<\/sub><\/p>\n

        \t\t\t\t\t\t\t\u00a0<\/td>\n

        		6\u00bd, 5, 3\u00bd, 2 \u2026.
        \n-2, –3<\/sup>\/4<\/sub>, \u00bd, 1 \u00be
        \n \u00a0\u00a0\u00a0\u00a0
        \n T2<\/sub> \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 T3<\/sub>
        \n\t\t\t\t\t\t \u00a0\u00a0\u00a0\u00a0a \u00a0\u00a0\u00a0\u00a0a + d <\/td>\n        		\"\"\/\"\"\/ -1\u00bd \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0
        \n 1\u00bc \u00a0\u00a0\u00a0\u00a0
        \n \u00a0\u00a0\u00a0\u00a0T4<\/sub> \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 T5 <\/sub>
        \n\t\t\t\t\t\t \u00a0\u00a0\u00a0\u00a0a + 2d a + 3d <\/td>\n        		 6\u00bd
        \n -2
        \n \u00a0\u00a0\u00a0\u00a0a + 4d <\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

        Ex: \u00a0\u00a0\u00a0\u00a0A.P \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 d (common difference) \u00a0\u00a0\u00a0\u00a0a (first term) \u00a0\u00a0\u00a0\u00a0 <\/p>\n

        \u00a0So for any A.P, the nth term (Tn<\/sub> = Un<\/sub>) is given by
        \n \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0Tn<\/sub> =Un<\/sub>= a + (n \u2013 1) d. Tn<\/sub>= Un<\/sub>= nth term \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 a = first term
        \n \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 d = common difference
        \n \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 n = no of terms <\/p>\n

        Examples \u00a0\u00a0\u00a0\u00a0
        \n\t<\/h4>\n

        1. What is the 10th<\/sup> term of the sequence 10, 6, 2, -4 \u2026.. 2. Find the term of the A.P 3\u00bd, 7, 10\u00bd \u2026.. Which is 77.
        \n3. The fist term of an A.P is 3 and the 8th<\/sup> term is 31. Find the common difference. <\/p>\n

        \u00a0<\/p>\n

        Solution
        \n<\/h4>\n
          \n
        1. The A.P = 10, 6, 2, -4 \u00a0\u00a0\u00a0\u00a0a = 10, d = 6 \u2013 10 = – 4, n = 10\n<\/li>\n<\/ol>\n

          \u00a0\u00a0\u00a0\u00a0Tn<\/sub> = a + (n \u2013 1) d
          \n \u00a0\u00a0\u00a0\u00a0T10<\/sub> = 10 + (10 \u2013 1) (-4) \u00a0\u00a0\u00a0\u00a0T10<\/sub> = 10 +9(-4) = 10 \u2013 36 \u00a0\u00a0\u00a0\u00a0T10<\/sub> = -26. <\/p>\n

            \n
          1. A.P = 3 \u00bd, 7, 10\u00bd \u2026\u2026\u2026\u2026\u2026\u2026 77 a = 3\u00bd, d = 7 \u2013 3\u00bd = 3\u00bd, n =? Tn<\/sub> = 77\n<\/li>\n<\/ol>\n

            \u00a0\u00a0\u00a0\u00a0Tn<\/sub> = a + (n-1)d
            \n \u00a0\u00a0\u00a0\u00a077 = 3\u00bd + (n-1)3\u00bd
            \n 77 = 3\u00bd + 3\u00bdn – 3\u00bd
            \n \u00a0\u00a0\u00a0\u00a077 = 3\u00bd n
            \n \u00a0\u00a0\u00a0\u00a0n = 77\/3\u00bd = 77\/7\/2 \u00a0\u00a0\u00a0\u00a0n = 77<\/sup> x 2<\/sup>\/7<\/sub> = 22 <\/p>\n

            \u00a0(3) \u00a0\u00a0\u00a0\u00a0a = 3, T8<\/sub> = 31, d = ? n = 8
            \nTn<\/sub> = a + (n-1) d
            \n 31 = 3 + (8-1) d
            \n 31 \u2013 3 = 7d
            \n \u00a0\u00a0\u00a0\u00a0d = 28<\/sup>\/7<\/sub> = 4 <\/p>\n

            Evaluation
            \n\t<\/h4>\n
              \n
            1. Find the 15th<\/sup> term of the A.P 5, 2, -1, -4 \u2026\u2026\u2026\u2026\n<\/li>\n
            2. Find the term of the A.P 1, 6, 11, 16\u2026. which is 66.\n<\/li>\n<\/ol>\n

              \u00a0<\/p>\n

              Arithmetic Mean
              \n\t<\/h4>\n

              If a, b, c are three consecutive terms of an A.P, then the common difference, d, equals b \u2013 a = c \u2013 b = common difference. b + b = a +c 2b = a + c
              \nb = \u00bd(a +c) <\/p>\n

              \u00a0<\/p>\n

              Examples
              \n<\/h5>\n
                \n
              1. Insert four arithmetic means between -5 and 10.\n<\/li>\n
              2. The 8th<\/sup> term of a linear sequence is 18 and the 12th<\/sup> term is 26. Find the first term, the common difference and the 20th<\/sup> term.\n<\/li>\n<\/ol>\n
                \n\n\n\n
                \n\u00a0\"\"\/Solution <\/strong>
                \n\t\t\t\t\t\t(i) \u00a0\u00a0\u00a0\u00a0Let the sequence be -5, a, b, c, d, 10. \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0a = -5, T6 = 10, n =6. \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0Tn<\/sub> = a + (n-1) d
                \n \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a010 = -5 + (6 \u2013 1) d
                \n \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a015 = 5d
                \n \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0d = 15<\/sup>\/5<\/sub> = 3
                \n \u00a0\u00a0\u00a0\u00a0a = -5 + 3 = -2 \u00a0\u00a0\u00a0\u00a0b = -2 + 3 = 1 \u00a0\u00a0\u00a0\u00a0c = 1 + 3 = 4 \u00a0\u00a0\u00a0\u00a0d = 4 + 3 = 7
                \nThe numbers will be -5, -2, 1, 4, 7, 10. <\/p>\n

                \u00a0
                \n\u00a0 \u00a0<\/td>\n

                		\n\u00a0
                \n\u00a0(ii) \u00a0\u00a0\u00a0\u00a0T8<\/sub> = a + 7d = 18, T12<\/sub> = a +11d = 26 a + 7d = 18 \u2026\u2026\u2026\u2026\u2026\u2026.. (i) \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0a + 11d =26 \u2026\u2026\u2026\u2026\u2026\u2026.. (ii)
                \n\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0Subtract (i) from (ii)
                \n \u00a0\u00a0\u00a0\u00a04d = 8 \u00a0\u00a0\u00a0\u00a0d = 2
                \n \u00a0\u00a0\u00a0\u00a0Substitute for d = 2 in (i) \u00a0\u00a0\u00a0\u00a0a + 7 (2) = 18
                \n \u00a0\u00a0\u00a0\u00a0a = 18 \u2013 14 \u00a0\u00a0\u00a0\u00a0a = 4
                \n\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0T20<\/sub> = a + (n \u2013 1) d = a + 19d
                \n\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0T20<\/sub> = 4 + (20 \u2013 1) 2
                \n\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 = 4 + 19 x 2
                \n\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0T20<\/sub> = 42<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

                Evaluation
                \n\t<\/h4>\n
                  \n
                1. Given that 4, p, q, 13 are consecutive terms of an A.P, find the values of p and q.\n<\/li>\n
                2. The sum of the 4th<\/sup> and 6th<\/sup> terms of an A.P is 42. The sum of the 3rd<\/sup> and 9th<\/sup> terms of the progression is 52. Find the first term, the common difference and the twentieth term of the progression.\n<\/li>\n<\/ol>\n

                  \u00a0<\/p>\n

                  Sum of terms in an A.P
                  \n\t<\/h4>\n

                  To find an expression for the sum of n terms of a linear sequence, Let Sn<\/sub> be the sum, then
                  \n\u00a0Sn<\/sub> = a + (a + d) + (a + 2d) + \u2026\u2026. + Tn<\/sub> \u2026\u2026\u2026\u2026.. (i) Also
                  \nSn<\/sub> = Tn<\/sub>+ (Tn<\/sub>– d) + (Tn<\/sub>– 2d) + \u2026\u2026\u2026 a \u2026\u2026\u2026. (ii)
                  \nAdding (1) and (2)
                  \n2Sn <\/sub>= (a + Tn<\/sub>) + (a + Tn<\/sub>) + (a + Tn<\/sub>) + \u2026\u2026\u2026\u2026 + (a + Tn<\/sub>)
                  \n\uf05c2Sn<\/sub> = n (a + Tn<\/sub>)
                  \n\uf05cSn<\/sub> = n<\/sup>\/2<\/sub> (a + Tn<\/sub>)
                  \nBut Tn<\/sub> = a + (n-1) d
                  \nSn<\/sub> = n<\/sup>\/2<\/sub> (2a + (n-1) d) <\/p>\n

                  \u00a0<\/p>\n

                  Examples<\/p>\n<\/h5>\n
                    \n
                  1. Find the sum of the first 25 terms of the A.P 3, 10, 17 \u2026\u2026\u2026.\n<\/li>\n
                  2. Find the sum of the first eight terms of a linear sequence whose first term is 6 and last term is 46.\n<\/li>\n
                  3. The sum of the first ten terms of an arithmetic progression is 255. Find the sum of the next twenty term of the A.P if the sum of the first twenty terms is 1010.\n<\/li>\n<\/ol>\n

                    \u00a0<\/p>\n

                    Solution
                    \n<\/h5>\n
                      \n
                    1. \"\"\/A.P = 3, 10, 17 \u2026\u2026\u2026\u2026..\n<\/li>\n<\/ol>\n

                      \u00a0\u00a0\u00a0\u00a0a = 3, d = 7, n = 25 \u00a0\u00a0\u00a0\u00a0 2a + 9d = 51 \u2026\u2026\u2026\u2026\u2026.(i) \u00a0\u00a0\u00a0\u00a0Sn<\/sub> = n<\/sup>\/2<\/sub> (2a + (n-1) d) \u00a0\u00a0\u00a0\u00a0 2a +19d = 101 \u2026\u2026\u2026\u2026\u2026\u2026 (ii)
                      \n \u00a0\u00a0\u00a0\u00a0 = 25<\/sup>\/2<\/sub> (2 x3 + (25 \u2013 1) 7) \u00a0\u00a0\u00a0\u00a0 Subtract (i) from (ii)
                      \n \u00a0\u00a0\u00a0\u00a0Sn<\/sub> = 25<\/sup>\/2 <\/sub>(6 +24 x 7) \u00a0\u00a0\u00a0\u00a0 10d = 50
                      \n \u00a0\u00a0\u00a0\u00a0S25<\/sub> =25<\/sup>\/2<\/sub> x174 = 2175 \u00a0\u00a0\u00a0\u00a0 d = 5
                      \n \u00a0\u00a0\u00a0\u00a0 Substitute for d = 5 in (i) <\/p>\n

                        \n
                      1. A.P , a = 6, Tn<\/sub> = 46, n = 8 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a02a + 9 x 5 = 51\n<\/li>\n<\/ol>\n

                        \u00a0\u00a0\u00a0\u00a0Sn<\/sub> = n<\/sup>\/2 <\/sub>(a + Tn<\/sub>) \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a02a = 51 \u2013 45
                        \n \u00a0\u00a0\u00a0\u00a0 = 8<\/sup>\/2 <\/sub>(6 + 46) \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a02a = 6
                        \n \u00a0\u00a0\u00a0\u00a0Sn = 4 (52) = 208. \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0a = 3
                        \n \u00a0\u00a0\u00a0\u00a0Sum of the next 20 terms = S30<\/sub> \u2013 S10<\/sub><\/p>\n

                          \n
                        1. S10<\/sub> = 10<\/sup>\/2<\/sub> (2a + (10 \u2013 1) d) = 255 \u00a0\u00a0\u00a0\u00a0S30<\/sub> = 30<\/sup>\/2<\/sub> (2 x 3 + (30 -1) 5) \u00a0\u00a0\u00a0\u00a0\n<\/li>\n<\/ol>\n

                          \u00a0\u00a0\u00a0\u00a0S20<\/sub> = 20<\/sup>\/2<\/sub> (2a + (20 \u2013 1) d) = 1010 \u00a0\u00a0\u00a0\u00a0 = 15 (6 + 29 x 5)
                          \n \u00a0\u00a0\u00a0\u00a0 5 (2a + 9d) = 255 \u00a0\u00a0\u00a0\u00a0 S30<\/sub> = 2265
                          \n \u00a0\u00a0\u00a0\u00a0 10 (2a + 19d) = 1010 \u00a0\u00a0\u00a0\u00a0S30<\/sub> \u2013 S10 <\/sub>= 2265 \u2013 255
                          \n \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 = 2010 <\/p>\n

                          \u00a0Evaluation<\/strong>:
                          \n\tThe sum of the first ten term of a linear sequence is -60 and the sum of the first fifteen term of the sequence is -165. Find the 18th<\/sup> term of the sequence. <\/p>\n

                          \u00a0<\/p>\n

                          General Evaluation
                          \n\t<\/h4>\n
                            \n
                          1. The sum of the first four terms of a linear sequence (A.P) is 26 and that of the next four terms is 74.\n<\/li>\n<\/ol>\n

                            Find the values of (i) the first term (ii) the common difference. <\/p>\n

                              \n
                            1. Calculate the (i) common difference (ii) the 20th<\/sup> term of the arithmetic progression; 100, 96, 92, 88, 86…\n<\/li>\n
                            2. Solve the equation: log4<\/sub>(x2<\/sup> + 6x + 11) = \u00bd\n<\/li>\n
                            3. Express 1 in the formm\u221a5 + n\u221a3<\/em><\/strong> where m and n are rational numbers\n<\/li>\n<\/ol>\n

                              3\u221a5 + 5\u221a3 <\/p>\n

                              \u00a0Reading Assignment<\/strong> \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0: Further Mathematics Project Book 1(New third edition).Chapter 28 -33 & 36 \u2013 37 <\/em><\/p>\n

                              \u00a0<\/p>\n

                              Weekend Assignment
                              \n\t<\/h4>\n
                                \n
                              1. Find T9<\/sub> of the sequence -1, 2, 5, 8 \u2026\u2026\u2026\u2026\u2026. A. 21 B. 22 C. 23 D. 24\n<\/li>\n
                              2. The 10th<\/sup> term of an A.P is 68 and the common difference is 7, find the first term of the sequence. A. 3 \u00a0\u00a0\u00a0\u00a0B. 5 \u00a0\u00a0\u00a0\u00a0 C. 7 D. 9\n<\/li>\n
                              3. Find the sum of the first twelve term of the sequence 2, 5, 8, 11… A. 202 B. 212 C.222 D. 232\n<\/li>\n
                              4. \n
                                What is the general term of the sequence 31, 26, 21, 16, 11\u2026\u2026\u2026\u2026\n<\/div>\n
                                  \n
                                1. 1 + 4n B. 3 x 2n-1 <\/sup>C. 36 \u2013 5n D. 5(\u00bd)n-2 <\/sup>\n\t\t\t\t<\/li>\n<\/ol>\n<\/li>\n
                                2. \n
                                  Find the sum of n terms of the A.P 3 + 6 +9 + 12 + \u2026\u2026\u2026.\n<\/div>\n
                                    \n
                                  1. 3n (n+1) B. 5n + 3<\/sup>\/2 <\/sub>(n+1) C. n + 3(n-1) D. 2n + n (n-3)\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                                    2 2 3 <\/p>\n

                                    Theory
                                    \n<\/h5>\n
                                      \n
                                    1. The first three terms of an A.P are x, 2x+1, 4x+1, find x and the sum of the first 18 terms.\n<\/li>\n<\/ol>\n

                                      The sum of the first twenty \u2013one terms of an A.P is 28, and the sum of the first twenty-eight terms is 21. Find which terms of the sequence is o and also the sum of the term proceeding it.<\/p>\n","protected":false},"excerpt":{"rendered":"

                                      \u00a0\u00a0\u00a0\u00a0 SECOND TERM E-LEARNING NOTE \u00a0 SUBJECT: FURTHER MATHEMATICS \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 CLASS: SS1 SCHEME…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,188],"tags":[],"class_list":["post-2164","post","type-post","status-publish","format-standard","hentry","category-posts","category-second-term-ss1-further-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2164","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=2164"}],"version-history":[{"count":2,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2164\/revisions"}],"predecessor-version":[{"id":2183,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2164\/revisions\/2183"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=2164"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=2164"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=2164"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}