Divisible by 2?

Number Theory Level 2

Suppose you have an arithmetic progression, with a term length of 4. This progression will be a subset of the set of all real integers, such as (22, 23, 24, 25) or (79, 81, 83, 85). Is it true that the sum of ANY of these progressions will be a multiple of the integer 2? Try to provide a proof or a counter-example in the solution discussion.

No Yes Insufficient information

2 solutions

Marta Reece
Jun 1, 2018

The sum of the four terms of the progression can be expressed as: $S=a+(a+b)+(a+2b)+(a+3b)=4a+6b$ where $a$ and $b$ are integers. Since $4$ and $6$ are both even, $S$ is even.

Excellent proof!

Brody Burkett - 3 years ago

Ram Mohith
Jun 1, 2018

Let the four terms be $a - 2d, a - d, a + d, a + 2d$ .

Sum = $a - 2d + a - d + a + d + a + 2d$

$Sum = 4a$

As, 4 is a multiple of 2 the sum will be always a multiple of 2.

This isn't a arithmetic progression.One of the difference is 2d,not d.

X X - 3 years ago

d is the common difference.

Ex : if 2 is the common difference with a = 4

The four terms are : 4 - 2(2) , 4 - 1(2) , 4 + 1(2) , 4 + 2(2)

So, the terms are 0, 2, 6, 8 which are in A.P

Ram Mohith - 3 years ago

The terms you said are 0,2,6,8, not 0,2,4,6,8

X X - 3 years ago

@X X – Actually the example which I took was wrong.

Ram Mohith - 3 years ago

Divisible by 2?

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