Four numbers are in GP, if the product of the extremes is 243 & sum of the middle two is 36,find the numbers
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The four numbers would be
a, ar, ar^2, and ar^3
a(ar^3) = 243
(a^2)(r^3) = 243 or ar(ar^2) = 243
ar + ar^2 = 36
ar(1 + r) = 36
divide ar(ar^2) = 243 by ar(1+r)=36
ar^2/(1+r) = 27/4
4ar^2 = 27 + 27r
a = (27+27r)/(4r^2) = 27(1+r)/(4r^2)
plug into: a^2 r^3 = 243
[ 27(1+r)/(4r^2) ]^2 r^3 = 243
729(1+r)^2/16r^4 (r^3) = 243
3(1+r)^2 /(16r) = 1
3(1+r)^2 = 16r
3 + 6r + 3r^2 = 16r
3r^2 - 10r + 3 = 0
(3r -1)(r - 3) = 0
r = 3 or r = 1/3
if r = 3, a = 27(4)/(36) = 3
the four terms are : 3 ,9,27,81
a, ar, ar^2, and ar^3
a(ar^3) = 243
(a^2)(r^3) = 243 or ar(ar^2) = 243
ar + ar^2 = 36
ar(1 + r) = 36
divide ar(ar^2) = 243 by ar(1+r)=36
ar^2/(1+r) = 27/4
4ar^2 = 27 + 27r
a = (27+27r)/(4r^2) = 27(1+r)/(4r^2)
plug into: a^2 r^3 = 243
[ 27(1+r)/(4r^2) ]^2 r^3 = 243
729(1+r)^2/16r^4 (r^3) = 243
3(1+r)^2 /(16r) = 1
3(1+r)^2 = 16r
3 + 6r + 3r^2 = 16r
3r^2 - 10r + 3 = 0
(3r -1)(r - 3) = 0
r = 3 or r = 1/3
if r = 3, a = 27(4)/(36) = 3
the four terms are : 3 ,9,27,81
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