Sum=21=A + Ar + Ar^2
sumsquares: 189=A^2+(Ar)^2 + (Ar)^4
21=A(1+r+r^2)
189=A^2(1+r^2+r^4)
squaring the first equation:
21^2=A^2 (1+r+r^2)^2
441=A^2 ( (1 +2r+3r^2+2r^3+r^4)
441=A^2 (1+r^2+r^4)+A^2(2r+2r^2+2r^3)
441=189+A^2*2r(1+r+r^2)
441=189+A^2*2r(21/A)
441=189+42A
solve for A. Now go back to the first equation and solve for r.
The su three numbers in g.p. is 21 and the sum of their squares is 189 find the numbers.
2 answers
A(1+r+r^2) = 21
A(r^3-1)/(r-1) = 21
A^2 (r^3-1)^2/(r-1)^2 = 441
A^2 (1+r^2+r^4) = 189
A^2 (r^6-1)/(r^2-1) = 441
Now divide and you get to cancel a lot of factors, winding up with
(r^2+r+1)/(r^2-r+1) = 441/189
cross-multiply and clean things up, and you end with
2r^2 - 5r + 2 = 0
(2r-1)(r-2) = 0
r = 2 or 1/2
A(r^3-1)/(r-1) = 21
A(7) = 21
A = 3
or
A(-7/8)/(-1/2) = 21
A(7/4) = 21
A = 12
check:
3+6+12=21
9+36+144=189
12+6+3=21
144+36+9=189
A(r^3-1)/(r-1) = 21
A^2 (r^3-1)^2/(r-1)^2 = 441
A^2 (1+r^2+r^4) = 189
A^2 (r^6-1)/(r^2-1) = 441
Now divide and you get to cancel a lot of factors, winding up with
(r^2+r+1)/(r^2-r+1) = 441/189
cross-multiply and clean things up, and you end with
2r^2 - 5r + 2 = 0
(2r-1)(r-2) = 0
r = 2 or 1/2
A(r^3-1)/(r-1) = 21
A(7) = 21
A = 3
or
A(-7/8)/(-1/2) = 21
A(7/4) = 21
A = 12
check:
3+6+12=21
9+36+144=189
12+6+3=21
144+36+9=189
2 answers