Week 1 – SS1 Second Term Further Mathematics Notes

    
SECOND TERM E-LEARNING NOTE

SUBJECT: FURTHER MATHEMATICS                CLASS: SS1 SCHEME OF WORK

REFERECES

• FutherMaths Project 1 and 2 by TuttuhAdegun (main text). ➢ Additional Mathematics by Godman
• Further Mathematics by E. Egbe et al.

WEEK ONE                          DATE…………… TOPIC: SEQUENCE & SERIES CONTENT

• Sequence and series
• Arithmetic Progression (AP)
• Arithmetic Mean
• Sum of terms in an AP

Sequence & Series

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.

 Consider the sequence 1, 4, 7, 10, 13 …., 1 is the first term,(T₁) 4 is the second term(T₂), 7 is the third term (T₃). 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

The nth term of a Sequence

The nth term of a sequence whose rule is stated may be represented by T_nso that T₁, T₂, T₃etc represent the first term, second term, third term … etc respectively. Consider the sequence 5, 9, 13, 17, 21 ……..
T₁ = 5 + 4(0)
T₂ = 5 + 4(1)
T₃ = 5 + 4 (2)

 T₄ = 5 + 4 (3)
T_n = 5 + 4 (n – 1)
T_n = 5 + 4n – 4 = 4n +1
when n = 30
T₃₀ = 4(30) + 1 T₃₀ = 121

 
 Find the nth term of these sequences:

1. 3, 5, 7, 9 …… 2n + 1
2. 0, 1, 4, 9 ……… (n -1)²
3. 1/3, 3/4, 1, 7/6 ………………2n – 1
                      n + 2

Examples

Write down the first four terms of the sequence whose general term is given by:
(i) T_n = n+1 (ii) T_n = 5 x (¹/₂)^n-2
3n +2
         T₄ = 4 + 1 = ⁵/₁₄
Solution      i.     T_n = n+1     
     3n + 2     (ii) T_n = 5 x (1/2)^n-2
    T₁ = 1 + 1 = ²/₅     T₁ = 5 x (1/2)^1-2 = 5(¹/₂)^-1 = 5(2^-1)^-1 = 5 x 2 = 10
     3(1) + 2     T₂ = 5 x (1/2)^2-2 = 5(¹/₂)⁰ = 5 x 1 = 5
    T₂ = 2 + 1 = ³/₈     T₃ = 5 x (1/2)^3-2 = 5 x (¹/₂) = ⁵/₂
     3(2) +2     T₄ = 5 x (1/2)^4-2 = 5(¹/₂)² = ⁵/₄
    T₃ = 3 + 1 = ⁴/₁₁     The sequence is 10, 5, ⁵/₂, ⁵/₄ ………
     3(3) + 2     
3(4) + 2
The sequence is ²/₅, ³/₈, ⁴/₁₁, ⁵/₁₄ ……..

Evaluation

Find the first term of the sequence whose general term is given by
(i) 50 – (½)ⁿ (ii) 2 + ³/₂⁽ⁿ⁺¹⁾

Arithmetic Progression (A.P) or Linear Sequence

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.

T1

6½, 5, 3½, 2 …. -2, –3/4, ½, 1 ¾      T2           T3     a     a + d

-1½           1¼          T4           T5     a + 2d a + 3d

6½ -2     a + 4d

Ex:     A.P                d (common difference)     a (first term)     

 So for any A.P, the nth term (T_n = U_n) is given by
         T_n =U_n= a + (n – 1) d. T_n= U_n= nth term                               a = first term
                         d = common difference
                              n = no of terms

Examples     

1. What is the 10^th term of the sequence 10, 6, 2, -4 ….. 2. Find the term of the A.P 3½, 7, 10½ ….. Which is 77.
3. The fist term of an A.P is 3 and the 8^th term is 31. Find the common difference.

Solution

1. The A.P = 10, 6, 2, -4     a = 10, d = 6 – 10 = – 4, n = 10

    T_n = a + (n – 1) d
    T₁₀ = 10 + (10 – 1) (-4)     T₁₀ = 10 +9(-4) = 10 – 36     T₁₀ = -26.

1. A.P = 3 ½, 7, 10½ ……………… 77 a = 3½, d = 7 – 3½ = 3½, n =? T_n = 77

    T_n = a + (n-1)d
    77 = 3½ + (n-1)3½
77 = 3½ + 3½n – 3½
    77 = 3½ n
    n = 77/3½ = 77/7/2     n = ⁷⁷ x ²/₇ = 22

 (3)     a = 3, T₈ = 31, d = ? n = 8
T_n = a + (n-1) d
31 = 3 + (8-1) d
31 – 3 = 7d
    d = ²⁸/₇ = 4

Evaluation

1. Find the 15^th term of the A.P 5, 2, -1, -4 …………
2. Find the term of the A.P 1, 6, 11, 16…. which is 66.

Arithmetic Mean

If a, b, c are three consecutive terms of an A.P, then the common difference, d, equals b – a = c – b = common difference. b + b = a +c 2b = a + c
b = ½(a +c)

Examples

1. Insert four arithmetic means between -5 and 10.
2. The 8^th term of a linear sequence is 18 and the 12^th term is 26. Find the first term, the common difference and the 20^th term.

Solution (i)     Let the sequence be -5, a, b, c, d, 10.          a = -5, T6 = 10, n =6.          Tn = a + (n-1) d          10 = -5 + (6 – 1) d          15 = 5d          d = 15/5 = 3     a = -5 + 3 = -2     b = -2 + 3 = 1     c = 1 + 3 = 4     d = 4 + 3 = 7 The numbers will be -5, -2, 1, 4, 7, 10.

(ii)     T8 = a + 7d = 18, T12 = a +11d = 26 a + 7d = 18 ……………….. (i)          a + 11d =26 ……………….. (ii)          Subtract (i) from (ii)     4d = 8     d = 2     Substitute for d = 2 in (i)     a + 7 (2) = 18     a = 18 – 14     a = 4          T20 = a + (n – 1) d = a + 19d          T20 = 4 + (20 – 1) 2           = 4 + 19 x 2          T20 = 42

Evaluation

1. Given that 4, p, q, 13 are consecutive terms of an A.P, find the values of p and q.
2. The sum of the 4^th and 6^th terms of an A.P is 42. The sum of the 3^rd and 9^th terms of the progression is 52. Find the first term, the common difference and the twentieth term of the progression.

Sum of terms in an A.P

To find an expression for the sum of n terms of a linear sequence, Let S_n be the sum, then
 S_n = a + (a + d) + (a + 2d) + ……. + T_n ………….. (i) Also
S_n = T_n+ (T_n– d) + (T_n– 2d) + ……… a ………. (ii)
Adding (1) and (2)
2S_n = (a + T_n) + (a + T_n) + (a + T_n) + ………… + (a + T_n)
2S_n = n (a + T_n)
S_n = ⁿ/₂ (a + T_n)
But T_n = a + (n-1) d
S_n = ⁿ/₂ (2a + (n-1) d)

Examples

1. Find the sum of the first 25 terms of the A.P 3, 10, 17 ……….
2. Find the sum of the first eight terms of a linear sequence whose first term is 6 and last term is 46.
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.

Solution

1. A.P = 3, 10, 17 …………..

    a = 3, d = 7, n = 25      2a + 9d = 51 …………….(i)     S_n = ⁿ/₂ (2a + (n-1) d)      2a +19d = 101 ……………… (ii)
     = ²⁵/₂ (2 x3 + (25 – 1) 7)      Subtract (i) from (ii)
    S_n = ²⁵/₂ (6 +24 x 7)      10d = 50
    S₂₅ =²⁵/₂ x174 = 2175      d = 5
     Substitute for d = 5 in (i)

1. A.P , a = 6, T_n = 46, n = 8          2a + 9 x 5 = 51

    S_n = ⁿ/₂ (a + T_n)          2a = 51 – 45
     = ⁸/₂ (6 + 46)          2a = 6
    Sn = 4 (52) = 208.          a = 3
    Sum of the next 20 terms = S₃₀ – S₁₀

1. S₁₀ = ¹⁰/₂ (2a + (10 – 1) d) = 255     S₃₀ = ³⁰/₂ (2 x 3 + (30 -1) 5)     

    S₂₀ = ²⁰/₂ (2a + (20 – 1) d) = 1010      = 15 (6 + 29 x 5)
     5 (2a + 9d) = 255      S₃₀ = 2265
     10 (2a + 19d) = 1010     S₃₀ – S₁₀ = 2265 – 255
          = 2010

 Evaluation:
The 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 18^th term of the sequence.

General Evaluation

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.

Find the values of (i) the first term (ii) the common difference.

1. Calculate the (i) common difference (ii) the 20^th term of the arithmetic progression; 100, 96, 92, 88, 86…
2. Solve the equation: log₄(x² + 6x + 11) = ½
3. Express 1 in the formm√5 + n√3 where m and n are rational numbers

3√5 + 5√3

 Reading Assignment          : Further Mathematics Project Book 1(New third edition).Chapter 28 -33 & 36 – 37

Weekend Assignment

1. Find T₉ of the sequence -1, 2, 5, 8 ……………. A. 21 B. 22 C. 23 D. 24
2. The 10^th term of an A.P is 68 and the common difference is 7, find the first term of the sequence. A. 3     B. 5      C. 7 D. 9
3. Find the sum of the first twelve term of the sequence 2, 5, 8, 11… A. 202 B. 212 C.222 D. 232
4. What is the general term of the sequence 31, 26, 21, 16, 11…………
   1. 1 + 4n B. 3 x 2^n-1 C. 36 – 5n D. 5(½)^n-2
5. Find the sum of n terms of the A.P 3 + 6 +9 + 12 + ……….
   1. 3n (n+1) B. 5n + ³/₂ (n+1) C. n + 3(n-1) D. 2n + n (n-3)

2 2 3

Theory

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.

The sum of the first twenty –one 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.