Question

Five distinct positive integers are in a arithmetic progression with a positive common difference. their sum is 10020, then the smallest possible value of the last term is​

Answer 1

Answer:

2002...2003.....2004...2005.....2006...

answer is 2006

Answer 2

Answer:

Step-by-step explanation:

Let the five posiive integers in ap be

a-10d,a-5d,a,a+5d,a+10d

A/Q

a-10+a-5d+a+a+5d+a+10d=10020

5a=10020

a=2004

Again,

S5=10020

5/2[a+l]=10020

[a+l]=10020×2/5

a+l=2(2004)

l=2(2004)-2004

l=2004