Question

In a sequence of four positive numbers, the first three are in geometric progression and the last three are in arithmetic progression. The first number is 12 and the last number is 45/2 . The sum of the two middle numbers can be written as a/b where a and b are coprime positive integers. Find a+b .

Answer 1

\[\left\{a _{1},a _{2},a _{3},a _{4}\right\}\]
1st three are geometric:
\[=\left\{ 12,12f,12f ^{2},\frac{ 45 }{ 2 } \right\}\]
Last three are arithmetic
\[=\left\{ 12,12f,12f+d,\frac{ 45}{2} \right\}\]
solve\[12f+2d=45/2\]
and \[12f ^{2}=12f+d\]
Two equations, two unknowns. Quadratic gives two solutions, take the ones that make series positive
I get for solution (you'll need to double-check), \[a+b =139\].

Answer 2

correct

Answer 3

\[Let the terms be 12,12 r,12r ^{2},\frac{ 45 }{2 } \]
where r is the common ratio of first three terms.
beacuse last three terms are A.P
\[12r+\frac{ 45 }{ 2 }=2*12r ^{2}\]
\[24r+45=48r ^{2}\]
\[48r ^{2}-24r-45=0\]
\[r=\frac{ 24\pm \sqrt{24^{2}-4*48*-45} }{ 2*48 }\]
\[r=\frac{ 24\pm \sqrt{576+8640} }{96 }\]
\[r=\frac{ 24\pm \sqrt{\sqrt{9216}} }{ 96 }\]
\[r=\frac{ 24\pm96 }{96 }=\frac{ 120 }{ 96 },\frac{ -72 }{96 }=\frac{ 5 }{ 4 },\frac{ -3 }{4 }\]
terms are positive hence r is positive.
middle terms are 12*\frac{ 5 }{4 },12*\frac{ 25 }{ 16 }\]
\[15,\frac{ 75 }{4 }\]
\[15+\frac{ 75 }{4 }=\frac{ 60+75 }{4}=\frac{ 135 }{ 4 }\]
a=135,b=4,a+b=135+4=139