Question

the 15th term of an arithmetic sequence is 143 and the 31 term is 183.
4) Find U1 and d
3) Find U100

Answer 1

Answer:

let 1 st term is a and common difference is d

so, a+(15-1).d=143

=> a+14d=143

=> a=143-14d

again,

a+(31-1).d=183

=> a+30d=183

=> a=183-30d

143-14d=183-30d

=> -14d+30d=183-143

=> 16d=40

=> d=40/16=10/4=5/2

so, a=143-14.5/2=143-35=108

U1=108 and d=5/2

U100=a+(100-1).d

=143+99×5/2

= 143+495/2

=143+247.5

=390.5