arithmetic progression

Arithmetic and Geometric progressions.

arithmetic progression

in an arithmetic progression, the sum of the 1st 3 terms is 18, and the sum of the squares of these terms is 126. find the terms. how is the solution?

Posts: 1
Joined: Sun Jun 20, 2010 3:04 am
Location: Philippines

suppose a,b,c are the first three terms of the arithmetic progression
then:
a + b + c = 18
a2 + b2 + c2 = 126

a, b, c form an arithmetic progression so:
a2 = b + c

so we have a system of three equations and 3 variables
from the first and the third we have
a + a2 = 18 - so we know a
Math Tutor

Posts: 313
Joined: Sun Oct 09, 2005 11:37 am

solution

The solution is as follows:

A + B + C = 18

A + C = 2B

A^2 + B^2 + C^2 = 126
So in solving the first two equations we will get B= 6

then A + C = 12 OR C = (12 - A )

then substitute in the third equation we get

A^2 + 6^2 + ( 12 - A )^2 = 126

we get A =3 OR A =9

The A.P will be 3 , 6 , 9 OR 9, 6 ,3
sabah orou

Posts: 2
Joined: Wed Jun 23, 2010 12:31 am

sabah orou you are right.
the problem comes from the fact that I have written b2 = a + c
and
it is
2b = a + c
Math Tutor