Question

Five distinct positive integers are in arithmetic progression with a positive common difference, if their sum is 10020

Answer 1

Step-by-step explanation:

2002,2003,2004,2005,2006

Answer 2

The smallest possible value of last term is  $l=2004+4d$.

Step-by-step explanation:

Given : Five distinct positive integers are in arithmetic progression with a positive common difference, if their sum is 10020.

To find : The smallest possible value of last term is ?

Solution :

Let the numbers be 'a-2d, a-d, a, a+d , a+2d '

The sum of the numbers is 10020.

i.e. $a-2b+a-b+a+a+b+a+2b=10020$

$5a=10020$

$a=\frac{10020}{5}$

$a=2004$

The smallest possible value of last term is

$l=a+(n-1)d$

Here, n=5 and a=2004

$l=2004+(5-1)d$

$l=2004+4d$

#Learn more

The sum of first three terms of an AP is 45 and the sum of their squares is 693.If common difference is positive,then its fourth term is

1.27

2.18

3.21

4.24

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