1. Example 1:
Find the 10th term and the nth term for the sequence 7, 10, 13, … .
Solution:
Un=
U10 =

2. Example 2
Find the three numbers in an arithmetic progression whose sum is 24 and whose product is 480.
Solution
Let the three numbers be (a-d), a, (a+d) where d is the common difference.
(a-d)(a)(a+d)=480
(a-d) +a + (a+d) = 24
Substituting a =8, we get
3a = 24
a = 8

3. When d = 2, required numbers are 6, 8, 10

4. Example 3
The sum of the first eighteen terms of an arithmetic series is -45 and the eighteenth term is also -45.
Find the common difference and the sum of the first hundred terms.
Solution
(1)

5. (2)
Solving equation 1 and 2, we get

6. Example 4
Find the sum of the positive integers which are less than 500 and are not multiples of 11.
Solution
Sum of multiples of 11(SII)
Sum of integers from 1 to 499 (SI)
Required sum=

7. Required sum = SI-SII =
=113365

8. Example 5
The sum of the first 10 terms of an A.P. is 145 and the sum of the next 6 terms is 231
Find (i) the 31st term , and
(ii) the least number of terms required for the sum to exceed 2000.

9. Solution
(1)
(2)

10. (1)-(2):
From (2):
(ii)Find the least number of terms required for the sum to exceed 2000.
Let n be the least number of terms required for

11. Consider

12. For + - +
Since n must be a positive integer, n > 36.7
Hence, the least number of terms required is 37.

13. Example 6
Given that the fifth term of a geometric
progression is and the third term is 9.
Find the first term and the common ratio if all the terms in the G.P. are positive.
(1)
(2)

14. (1)(2):
Since all the terms in the G.P. are positive, r > 0
From (2):

15. Example 7
Three consecutive terms of a geometric
progression are and 81. Find the value of x. If 81 is the fifth term of the geometric progression, find the seventh term.

16. Given 81 is the fifth term, find the seventh term:
U5= ar4 = 81
U7 = ar6

17. Example 8
Find the sum of the first eight terms of the series
Soln:
G.P. with

19. Example 9
A geometric series has first term 1 and the common ratio r, where r 1, is positive. The sum of the first five terms is twice the sum of the terms from the 6th to 15th inclusive. Prove that

20. Solution
Given
Since and a = 1, we have

21. Let
So the equation becomes
Let f(x) =
Since f(1) = 2 – = 0
(x –1) is a factor of f(x).
Comparing coefficient of x:

22. Hence,
(r<0)
Since
and r > 0,

23. Sum to infinity, S (or ) =
The sum to infinity exists (the series converges or series is convergent) provided
Example 10
Determine whether the series given below converge. If they do, give their sum to infinity.
G.P. with >1
(a)
Does not converge

24. G.P. with
(b)
<1
Series converges

25. G.P. with
(c)
>1
Does not converge

26. Example 11
A geometric series has first term a and the common ratio
Show that the sum to infinity of the geometric progression is
Solution
(shown)

27. Example 12
A geometric series has first term a and common ratio r. S is the sum to infinity of the series, T is the sum to infinity of the even-numbered terms (i.e ) of the series. Given that S is four times the value of T, find the value of r.
common ratio r 2
Given S = 4T, we have