Arithmetic progression

Arithmetic and Geometric progressions.

Arithmetic progression

Postby Maria Joe Trinidad » Sun Jun 20, 2010 3:11 am

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?
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Postby Math Tutor » Sun Jun 20, 2010 11:48 am

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

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solution

Postby sabah orou » Wed Jun 23, 2010 1:10 am

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

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Postby Math Tutor » Wed Jun 23, 2010 4:09 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

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