If ${\sqrt 3} – {\sqrt 2}, 4- {\sqrt 6}, p{\sqrt 3} – q {\sqrt 2}$ form a geometric progression, find the values of p and q.
So I take the second term $4-{\sqrt 6} =( {\sqrt 3} – {\sqrt 2}) (r)$ , where r is the common ratio.
$4-{\sqrt 6} =( {\sqrt 3} – {\sqrt 2})( 2{\sqrt3} + {\sqrt2 })$
And found that the common ratio, r = $2{\sqrt3} + {\sqrt2 }$
To find the third term, I multiplied the second term with the common ratio.
$(4-{\sqrt 6})( 2{\sqrt3} + {\sqrt2 })= p{\sqrt 3} – q {\sqrt 2}$
$8{\sqrt 3} + 4{\sqrt2} – 6 {\sqrt 2} – 2{\sqrt 6} = p{\sqrt 3} – q {\sqrt 2}$
I am unable to proceed beyond this step.
Best Answer
You've made a computation error, perhaps? $$(4-\sqrt6)(2\sqrt3+\sqrt2)=8\sqrt3+4\sqrt2-6\sqrt2-2\sqrt3=6\sqrt3-2\sqrt2$$