Page 8 - Grab Me SPM Add Mathematics Form 4,5

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Example 17 Example 18
The sum of the first three terms of a geometric Given a geometric progression with T = 1 and r = 2.
1
progression with a common ratio – 2 is 63. Find Calculate the minimum value of n if the sum of the
the first term. 3 first n terms is more than 2 000.
Solution Solution
S = 63 S  2 000
n
3
a(1 – (– ) ) 1(2 – 1)  2 000
2
n
3
3 = 63 2 – 1
1 – (– ) 2  2 001
2
n
3 n log 2  log 2 001
35 a = 63( ) 10 10
5
log 2 001
10
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3
n 
log 2
27
5
a = 63( )( ) n  10.97 10
3
35
a = 81 Therefore, n = 11.
Another Method
Example 19
T + T + T = 63
3
1
2
a + ( – 2a ) ( )( – 2a ) = 63 The nth term of a geometric progression is
2
+ –
3 3 3 T = 4(4 1 – n ). Calculate
n
9a – 6a + 4a = 63 (a) the common ratio,
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a = 81 (b) the sum to infinity.
Ch 5_Grab Me SPM AddMaths F4.indd 60 13/05/2022 9:01 AM