a,b,c, are natural numbers and they are the terms of a G.P. If a+b+c is an even number, prove that a,b,c, are even numbers.

Expert Answers
hala718 eNotes educator| Certified Educator

a, b , c are terms in geometric progression:

a= a

b = a*r

c= a*r^2

==> a+b + c = a+ ar + ar^2

But a+ b + c is an evern number.

==> a+b + c = 2n (n=1,2,3....)

==> a + ar + ar^2 = 2n

==> Factor a:

==> a(1+r+r^2) = 2n = even

But 1+r + r^2 is an odd number

==> a should be an even number  because even*odd= even

==> a = even = 2n

b = a*r = 2nr = even

c= ar^2 = 2nr^2 = even

giorgiana1976 | Student

If a+b+c is an even number, we'll write:

a+b+c = 2k

Also, a,b,c are the terms of the geometric progression. We'll put q as the constant of the progression.

a = a

b = a*q

c = a*q^2

We'll re-write the sum:

a+b+c = a + a*q + a*q^2 = 2k

We'll factorize:

a(1+q+q^2) = 2k

But 1+q+q^2 is an odd number.

q+q^2 is an even number because q and q^2 have the same parity.

So, a = 2p

b = a*q = 2p*q

c = a*q^2 = 2pq^2

So, the numbers a,b,c are even numbers, too.