Page 15 - PRE-U STPM MATHEMATICS (T) TERM 1

Mathematics Term 1 STPM Chapter 2 Sequences and Series

This series is an arithmetic series with 1st term, a = ln 3 and common difference
d = ln 3, i.e.
a = d = ln 3.
10
∑ ln 3 = ln 3 + 2 ln 3 + 3 ln 3 + … + 10 ln 3
r
r = 1 = (1 + 2 + 3 + … + 10) ln 3
= 10 (1 + 10) ln 3
2
= 55 ln 3
n
and ∑ ln 3 = (1 + 2 + 3 + … + n) ln 3
r
r = 1 n
= (1 + n) ln 3.
2
2 We have to find the smallest value of n such that
n
(1 + n) ln 3  1000
2
n(1 + n)  2000
ln 3
n(1 + n)  1821
By using the calculator, we find that
if n = 42, n(1 + n) = 42 × 43 = 1806 , 1821
if n = 43, n(1 + n) = 43 × 44 = 1892  1821
n
Hence, the smallest value of the integer n such that ∑ ln 3  1000 is 43.
r
r = 1
Example 13

th
The n term of an arithmetic progression is 32 and its first term is 2. If the sum of the first n terms is 357,
th
find the value of n. If the smallest term in the series which exceeds 100 is the k term, find the value of k.
st
Solution: The 1 term, a = 2
th
The n term, a + (n – 1)d = 32 d = common difference
2 + (n – 1) d = 32
(n – 1) d = 30 ………… 
st
Sum of the 1 n terms is 357.
Thus n [2a + (n – 1) d] = 357
2
i.e. n (4 + 30) = 357
2 17n = 357
n = 21
Substitute n = 21 into ,
20d = 30
d = 3
2
th
The k term, u = a + (k – 1) d
k
= 2 + (k – 1) · 3
2
= 1 (3k + 1)
2
For 1 (3k + 1)  100
2
3k + 1  200
3k  199
k  66.3
Hence, the smallest term in the series which exceeds 100 is the 67 term, i.e. k = 67.
th

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02 STPM Math T T1.indd 102 3/28/18 4:21 PM