
An arithemtic progression (AP) is a sequence of terms in which any term minus its immediate preceding term gives a constant.
This constant is called the common difference.
Let a be the first term and d be the common difference of an AP.
The nth term of the AP is T_{n} = a + (n  1)d.
The sum of the first n terms is S_{n}  = ½n[2a + (n  1)d] 
= ½n[a + T_{n}]. 
A geometric progression (GP) is a sequence in which each term is obtained from the preceding one by multplying it by a constant.
This constant is called the common ratio.
Let a be the first term and r be the common ratio of a GP.
The nth term of the AP is T_{n} = ar^{n}^{  1}.
The sum of the first n terms is S_{n} = 
¾¾¾¾ r  1 
or 
¾¾¾¾ 1  r 
. 
The sum to infinity is S_{¥} = 
¾¾¾ 1  r 
provided that r < 1. 
Suppose we know the formula for general term T_{n} of a progression.
Compute the term T_{n + 1}.
Example 1: Find the sum of all integers between 1000 and 3000 inclusive which are not divisible by 7.
Solution:
\  Sum of all integers between 1000 and 3000 
= (2001/2)[1000 + 3000]  
= 4 002 000 
Numbers between 1000 and 3000 divisible by 7 are:
This is an A.P. with a = 1001 and d = 7.
\  Sum of all integers between 1000 and 3000 divisible by 7 
= (286/2)[1001 + 2996]  
= 571 571 
\  Sum of all integers between 1000 and 3000 which are not divisible by 7 
= 4 002 000  571 571  
= 3 430 429 
Example 2: An arithmetic series has first term 2000 and common difference 2.3. Calculate the value of the first negative term of the series, and the sum of all the positive terms.
Solution:
T_{n} = a + (n  1)d

<  0 
2000 + (n  1)(2.3)

<  0 
(n  1)(2.3)

<  2000 
n  1

>  2000/2.3 
n

>  870.56... 
The first negative term is the 871st term.
\ Value of first negative term  = 2000 + (871  1)(2.3) 
= 1 
Number of positive terms = 870.
\ Sum of all positive terms  = (870/2)[4000 + (870  1)(2.3)] 
= 1 741 131 
Example 3: The first two terms of a G.P. are 4 and 3. Write down an expression for the sum to n terms and evaluate the sum to infinity of this progression. Find the least value of n for the sum to n terms to be within 1% of the sum to infinity.
Solution:
Sum to n terms, S_{n} = 
¾¾¾¾¾¾ 1  (3/4) 

= 
¾¾¾¾¾¾ 7 

Sum to infinity, S_{¥} =



=  16/7 
S_{n} 
S_{¥}

<  1% S_{¥}  
(16/7)[1 
(3/4)^{n}]  16/7

<  (1/100)(16/7)  
(16/7)(3/4)^{n}

<  (1/100)(16/7)  
(3/4)^{n}

<  1/100  
n lg (3/4)

<  lg (1/100)  
n

> 

\ Least value of n = 17.