Page 108 - Mathematics_Form_3

Sequences and series

Example 4.21 364
−
If the sum of the first four terms of 3
a geometric progression is 80 and its Therefore, the sum of the first six terms
364
common ratio is 3. Find the first term. is − .
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Solution 3
Given S = 80, n = 4, r = 3. Thus,
4 Example 4.23
n
G ( r − 1)

S = 1 _
n r − 1 The n term of a geometric progression
th
Substituting the values into the formula is given by 24 (5) . Determine the;
n−1
gives, 4
G (3 − 1)

1
80 = _ (a) first five terms of the geometric
3 − 1 progression.
G (81 − 1)
= 1 _

3 − 1 (b) sum of the first 10 terms of the
40 G = 80 geometric progression.
1
G = 2 (c) general formula for the sum of first n
1
Therefore, the first term is 2. terms of the geometric progression.
(d) smallest value of n for which the
Example 4.22 sum, S > 90,000.
n
Find the sum of the first six terms of a Solution
geometric series − 81 − 27 − 9 − · · · (a) From G = G r , it implies that:
n−1
n 1
Solution The first term,
1 __
Given G = − 81, n = 6, r = . 1−1 0
1 3 G = 24 (5) = 24 (5) = 24
1
The formula for the sum of geometric The second term,
series is given by:
2−1
1
G = 24 (5) = 24 (5) = 120
2
n
G (1 − r ) The third term,

S = 1 _
n 1 − r
G = 24 (5) = 24 (5) = 600
3−1
2
Substituting the given values into the 3 Mathematics for Secondary Schools
formula gives, The fourth term,
G = 24 (5) = 24 (5) = 3,000
3
4−1
6 4
1 __
− 81 1 −
( ) )
(
3
S = ______________ The fifth term,
6 1 __
1 − G = 24 (5) = 24 (5) = 15,000 .
5−1
4
3 5
1
_
− 81 1 − Therefore, the first five terms are 24,
))
(
(
729
= _______________
2 __ 120,600, 3,000 and 15,000.

3
Student\s Book Form Three 101
18/09/2025 09:59:25
MATHEMATIC F3 SB.indd 101
MATHEMATIC F3 SB.indd 101 18/09/2025 09:59:25