Question

given three non zeros numbers ; x,y and zwhich are consecutive terms of a geometric progression ,and x + y + z =70.Further more 4x ,5y and 4z are consecutive terms of an arithmetic progression .Find the value of x, y and z

Answer 1

A geometric progression (GP) is one where each term is obtained by multiplying the previous term by some common factor, say r.
so, from the first statement, we can say that if x, y and z are consecutive terms of a GP, then:\[y = rx\tag{a}\]\[z = ry = r(rx)=r^2x\tag{b}\]
And we are given:\[x+y+z=70\]therefore:\[x+rx+r^2x=70\]therefore:\[x(1+r+r^2)=70\tag{c}\]
Next, an arithmetic progression (AP) is one where each term is obtained by adding a common difference, say d, to the previous term.
so, since you are told that 4x, 5y and 4z are consecutive terms of an AP, then:\[5y=4x+d\tag{d}\]\[4z=5y+d=(4x+d)+d=4x+2d\tag{e}\]Now lets substitute (a) into (d) to get:\[5rx=4x+d\tag{f}\]and lets substitute (b) into (e) to get:\[4r^2x=4x+2d\tag{g}\]Now lets eliminate d by doing 2 times (f) minus (g) to get:\[10rx-4r^2x=8x-4x=4x\]therefore:\[5rx-2r^2x=2x\]therefore:\[2r^2x-5rx+2x=0\]therefore:\[x(2r^2-5r+2)=0\implies 2r^2-5r+2=0\]therefore:\[(r-2)(r-0.5)=0\implies r=2\text{ or }r=0.5\]Now use these solutions to r to find x from equation (c) above, then find y and z from equations (a) and (b) above.

Answer 2

sure that`s great asneeer and through my readings i met the other solution.do u wish that i can share it with you ?

Answer 3

there should be two sets of solutions for x, y and z.
it would be great if you could list the ones you found so I can then confirm if they match my answers.

Answer 4

sure it`s as follows ;
x,y and z in G.P
x + y +z =70
4x,5y and 4z in A.P
from the ratio r=G2/G1=G3/G2
y^2 =xz.............................................................(i)
common difference d =A2-A1 =A3 -A2
5y -4x =4z -5y
10y=4x + 4z
10y=4(x +z)
5y/2=(x +z)........................................................(ii)
from;
x+y+z=70
y +(x+z) =70
but x+z =5y/2
y +5y/2=70......................................................(iii)
2y +5y=140
7y =140
y=20
from;
(y)^2 =xz
20^2=xz
400=xz...................................................(iv)
from 5y/2 =x +z
5(20)/2 =x +z
50 = x+z......................................(v)
z =400/x
50=x + (400/x)
x^2 -50x +400 =0
by the general formula
x =40 or 10
from z =400/x
z=10 or 40

therefore the value of y=20 and x =40 when z is 10 and x =10 when z =40
Thanks by John

Answer 5

yup - matches the answers I got as well - thanks for getting back to me.