• Given a sequence : 1,6,11,16
• The nth term of arithmetic progression is given by :
An = a + ( n-1 ) d ;
where a is the first term , a = 1
and d is the difference between two consecutive numbers = 6-1 = 5 ; or 11-6 = 5, therefore d = 5
1. The nth term will be :
An = 1 +(n-1) 5
= 1+ 5n -5
An = 5n-4
2. For the 100th term :
A(100) = 1+(100 -1) 5
= 1 + 500 -5
= 500 -4
A(100) = 496
• ,The 100th term is 496,
