Sequence And Series II SS1 Further Mathematics Lesson Note

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 the common ratio designated by the letter r.

Examples:

4, 8, 16, 32,      ……………. 8/4  = 2              

8, 4, 2, 1, ½,      ………….   4/8   = ½ 

3,-1, 11/3, -1/9 ………..       -1/3  = -1/3

For any G.P, the nth term is given by

Tn = arn-1

Tn = nth term

a = first term

r = common ratio

n = number of terms

Examples:      

1. Find the 9th term of the sequence G.P 2, -10, 50 ………….
2. Find the number of terms of the G.P 27, 81, 243 …………. 320
3. If 7, x, y, 189 are in G.P, find x and y

Solution

(1) G.P = 2, -10, 50 ………………

a = 2, r = -5, n = 9, T9 = ?

Tn = arn-1

T9 = 2(-5)9-1

= 2 x 390625

      T9 = 781250  

(2) G.P = 27, 81, 243 ………320

a = 27, r = 3, n =?, Tn = 320

Tn = arn-1

320 = 27(3)n-1

320 = 33(3)n-1

320 = 33+n-1

320 = 32+n

2+n = 20

n = 20 – 2 = 18

(3) The G.P = 7, x, y, 189.

a = 7, n = 4, T4 = 189

Tn = arn-1 = 189

7(r)4-1 = 189

r3 = 189/7   = 27 = 33

r = 3

T2 = x = ar = 7x 3 = 21

T3 = y ar2  = 7x3x3 = 63

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 means between 12 and 324.

(2) The 2nd term of an exponential sequence is 9 while the 4th 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

x = T2 = ar = 12 x 3 = 36

y = 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 

The 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 Sn = ar + ar2 + ar3 + ………….. arn…………. (ii)

 Subtracting (2) from (1)

Sn – rSn = a – arn

Sn (1- r) = a(1 – rn)

Sn = a(1 – rn)     if r ( 1

          1 – r

Sn = 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 the 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

2. G.P = 18, 6, 2 …….

a = 18, r = 6/18 = 1/3, n = 6, S6?

Sn = a (1 – rn)

                   1 – r

    = 18(1- (1/3)6)   = 18(1- 1/729)

                      1–1/32/3

S6   = 18 x 3 x 728

                      2 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