Week 2 – SS1 Second Term Further Mathematics Notes

WEEK TWO                                                  DATE…………… TOPIC: SEQUENCE AND SERIES (II) CONTENT

• Nth term of a G.P
• Geometric Mean
• Sum of n terms of a G.P
• Sum to infinity of a G.P

Nth term of a G.P

A Geometric Progression is a sequence generated by multiplying or dividing a preceding term by a constant number to get a term. This constant number is called common ratio designated by letter r.

 Examples:                r           a
4, 8, 16, 32, ……………..          ⁸/₄ = 2      4
8, 4, 2, 1, ½               ⁴/₈ = ½          8
3,-1, 1¹/₃, ^-1/₉               ^-1/₃ = ^-1/₃          3

 For any G.P, the nth term is given by          
Tn = arn-1
         
T_n = nth term (2)     G.P = 27, 81, 243 ………3²⁰ a = first term     a = 27, r = 3, n =?, T_n = 3²⁰ r = common ratio          T_n = arn-1
n = number of terms          3²⁰ = 27(3)^n-1
         320 = 33(3)n-1

Examples:          3²⁰ = 3^3+n-1

1.     Find the 9^th term of the sequence G.P 2, -10,          3²⁰ = 3²⁺ⁿ
50 ………….          2+n = 20

1. Find the number of term of the G.P 27, 81, 243          n = 20 – 2 = 18

…………. 3²⁰               

1. If 7, x, y, 189 are in G.P, find x and y     (3)     The G.P = 7, x, y, 189.

    a = 7, n = 4, T₄ = 189
    Solution     T_n = ar^n-1 = 189
(1)     G.P = 2, -10, 50 ………………     7(r)^4-1 = 189     a = 2, r = -5, n = 9, T₉ = ?          r³ = ¹⁸⁹/₇ = 27 = 3³
    T_n = ar^n-1     r = 3
    T₉ = 2(-5)^9-1               T₂ = x = ar = 7x 3 = 21
    = 2 x 390625     T₃ = y ar² = 7x3x3 = 63
T₉ = 781250

Evaluation

1. Find T₉ of the G.P 5, 2½, 1¼, ⁵/₈ …………..
2. If 3, p, q, 24 are consecutive term of an exponential sequence, find the values of p and q.

Geometric Mean

Suppose x, y, z are consecutive terms of a geometric progression, then the common ratio r can be written as:          r = y/x = z/y
         y/x = z/y           y2 = xz           y = xz     
    y is the geometric mean of x and z. Examples:

1. Insert two geometric mean between 12 and 324.
2. The 2^nd term of an exponential sequence is 9 while the 4^th term is 81. Find the common ratio and the first term
   of the G.P Solution

(1)     Let the G.P = 12, x, y, 324.     a = 12, T₄ = 324, n = 4     Tn = arn-1
    324 = 12(r)^4-1
    r3 = 324/12 = 27 = 33
    r = 3

1. = T² = ar = 12 x 3 = 36
2. = T₃ = ar² =12 x 3 x3 = 108 The geometric means are 36 and 108

 2) T₂ = 9, T₄ = 81
T₄ = ar³ = 81 ……………………(i)
T₂ = ar = 9 …………………… (ii)
Divide (i) by (ii) ar³ = 81/9 ar¹
r² = 9  r = + √ 9 = +√3 ar = 9 a(+3) = 9
a = 9 = + 3
+3
The first term = + 3, the common ratio = + 3
    

Evaluation

1. Insert two geometric mean between -3 and –⁸/_9.
2. The 4
       ^th term of a G.P in 75 and the 6^th term is 192. Find the common ratio and the first term of the G.P

Sum of n terms of a G.P

The sum of n terms of a G.P whose first term is a and whose common ratio is r is given by
Sn= a + ar + ar2 + ………………… arn-1 ……………..(i) r S_n = ar + ar² + ar³ + ………….. arⁿ…………. (ii) Subtracting (2) from (1)
S_n – rS_n = a – arⁿ
S_n (1- r) = a(1 – rⁿ)
S_n = a(1 – rⁿ) if r  1
1 – r
S_n = a(rⁿ– 1) if r  1 r – 1
Examples:

1. The third term of a G.P is 63 and the fifth term is 567. Find the sum of the first six terms of the progression.
2. Find the sum of first 6 terms of the G.P 18, 6, 2 …………

Solution

1. T₃ = 63, T₅ = 567

         T₅ = ar⁴ = 567 ……………. (i)
         T₃ = ar² = 63 ……………. (ii)          Divide (i) by (ii)
         ar⁴= 567
ar² 63          r² = 9          r = 3
         Substitute for r = 3 in (ii)
         a (3)² = 63
         a = ⁶³/₉ = 7
    S₆ = a (rⁿ – 1)
         r – 1
     = 7 (3⁶ – 1)
          3 – 1
     = 7(729 – 1)
2
    S₆ = 2548

1. G.P = 18, 6, 2 …….     a = 18, r = ⁶/₁₈ = ¹/₃, n = 6, S₆?

S_n = a (1 – rⁿ)

1. – r

       = 18(1- (¹/₃)⁶) = 18(1- ¹/₇₂₉)
1–¹/₃²/₃
    S₆ = 18 x 3 x 728

1. x 729

    S₆ = 26.9

Sum to Infinity

The sum of the n terms as n approaches infinity is called the sum to infinity of the series and is designated S Thus:
    S = a if r1
1-r
    S = a if r1
r-1
Examples:
Find the sum to infinity of the sequence 1, ¼, ¹/₁₆, ¹/₆₄.

Solution

    a = 1, r = ¼
    S = 1 = 1
1 – ¼ ¾
    S = ⁴/₃

Evaluation

1. The second and fourth terms of a G.P are 21 and 189. Find the sum of the first seven terms.
2. Find the sum to infinity of 1+¹/₃ + ¹/₉ + ¹/₂₇ …………

GENERAL EVALUATION

1. Find the (a)sum of the first 8 terms (b)sum to infinity of the series: -5 , 5/2, -5/4 , 5/8…….
2. The sum of the first two terms of a G.P is 2½ and the sum of the first four terms is 3¹¹/18. Find the G.P if r  0.
3. Solve the following exponential equations (a) 2^2x -6(2^x) + 8 = 0 (b) 2^2x+1 -5(2^x) + 2 = 0

 READING ASSIGNMENT: Further Mathematics Project Book 1(New third edition).Chapter 33-36 & 37-45
    

WEEKEND ASSIGNMENT

1. The sum to infinity of a G.P is 60. If the first term of the series 12, find its second term of the series 12, find its second term. A. 9.6 B. 6.9 C. 12.6 D. 8.6
2. A G.P has 6 terms. If the 3^rd and 4^th terms are 28 and -56 respectively, find the sum of the G. P.
   1. 471 B. -471 C. – 147 D. -741
3. Find the sum of the G.P 2 + 6 + 18 + 54 + …………1458. A. 8216 B. 6218 C.1682 D. 2186 4.     The 8^th term of a G.P is -7/32. Find its common ratio if its first term is 28.
   1. ½ B. -½ C. –²/₃ D. ³/₂

5.     Given the geometric progression 5, 10, 20, 40, 80 …………….. find its nth term.
    A. 2(5ⁿ⁺²) B. 5(2ⁿ⁺¹) C. 5(2^n-1) D. 2(5^n-1)

THEORY

1. The fifth term of a G.P is greater than the fourth term by 13½, and the fourth term is greater than the third by 9.

Find (i) the common ratio (ii) the first term

1. The sum of the first two terms of an exponential sequence is 135 and the sum of the third and the fourth terms is 60. Given that the common ratio is positive, calculate

(i)     the common (ii)     the limit of the sum of the first n terms as n becomes large
(iii)     the least number of terms for which the sum exceeds 240