Page 14 - PRE-U STPM MATHEMATICS (T) TERM 1

Mathematics Term 1 STPM Chapter 2 Sequences and Series

Example 10

Find the sum of the integers which can be divided by 6 exactly and which lie in between 50 and 150.
st
Solution: The 1 term divisible by 6 is 54.
The last term divisible by 6 is 144.
The required series is
S = 54 + 60 + 66 + … + 144
n
This is an arithmetic progression with a = 54 and d = 60 – 54 = 6.
The last term, l = a + (n – 1)d = 144
Hence, 54 + (n – 1)6 = 144
6n = 96 2
n = 16
Hence, the sum of this series is
S = 16 (a + l)
2
16
= 16 (54 + 144)
2
= 8 × 198
= 1584

Example 11

In an arithmetic progression, the sum of the first ten terms is 520 and the 7 term is twice the 3 term.
th
rd
Find the first term, a, and the common difference, d.
Solution: The sum of the first ten terms is
S = 10 (2a + 9d) = 520
10
2
2a + 9d = 104 ………… 
rd
th
The 7 term is a + 6d and the 3 term is a + 2d.
a + 6d = 2(a + 2d)
a = 2d ………… 
Substituting  into : 4d + 9d = 104
13d = 104
d = 8
a = 16
st
Hence, for the given arithmetic progression, the 1 term is 16 and the common
difference is 8.

Example 12
n
r
Find the sum of the first 10 terms of the series ∑ ln 3 , and the smallest value of n such that the sum of
the first n terms exceeds 1000. r = 1

r
Solution: By substituting r = 1, 2, 3, … into ln 3 , we have
n
2
r
3
∑ ln 3 = ln 3 + ln 3 + ln 3 + … + ln 3 n
r = 1
= ln 3 + 2 ln 3 + 3 ln 3 + … + n ln 3.
101

02 STPM Math T T1.indd 101 3/28/18 4:21 PM