The Five Positive Numbers

Algebra Level pending

Let a, b, c, d and e are real positive numbers, such that
{a, b, c, d, e} > 0 and $a\neq b\neq c\neq d\neq e$ .

Now consider following properties of {a, b, c, d, e}
1. a, b, c in A.P. (Arithmetic Progression)
2. b, c, d in G.P. (Geometric Progression)
3. c, d, e in A.P. (Arithmetic Progression)
4. $a^{2}$ + e = [ a $\times$ b ] + d
5. c + d + e = [ 2 $\times$ $a^{2}$ ]

If [ a + b + c + d + e ] is represented in terms of $\frac{p}{q}$ , where p and q are co-primes.
What is value of [ 3 $\times$ q ] - p .

The answer is 19.

1 solution

Abhimanyu Singh
Apr 11, 2014

$Property 1.\quad \{ a,b,c\} \quad in\quad AP\\ so,\quad \forall D>0\\ b=a+D\\ c=a+2D\\ Property 2.\quad \{ b,c,d\} \quad in\quad GP\\ so,\quad { c }^{ 2 }=bd\Rightarrow d=\left( \frac { { \left( a+2D \right) }^{ 2 } }{ \left( a+D \right) } \right) \\ Property 3.\quad \{ c,d,e\} \quad in\quad AP\\ so,\quad 2d=c+e\Rightarrow e=\left( \frac { \left( a+2D \right) \left( a+3D \right) }{ \left( a+D \right) } \right) \\ Property 4.\quad { a }^{ 2 }+e=ab+d\Rightarrow e-d=a(b-a)\Rightarrow e-d=aD\\ so,\quad \frac { D(a+2D) }{ a+D } =aD\Rightarrow D=\frac { a-{ a }^{ 2 } }{ a-2 } \\ Thus\boxed { b=\frac { a }{ 2-a } \\ c=\frac { { a }^{ 2 } }{ 2-a } \\ d=\frac { { a }^{ 3 } }{ 2-a } \\ e=\frac { { a }^{ 2 }(2a-1) }{ 2-a } } \\ Property 5.\quad c+d+e=2{ a }^{ 2 }\Rightarrow \frac { 3{ a }^{ 3 } }{ 2-a } =2{ a }^{ 2 }\Rightarrow a=\frac { 4 }{ 5 } \\ Thus\boxed { a=\frac { 4 }{ 5 } \\ b=\frac { 2 }{ 3 } \\ c=\frac { 8 }{ 15 } \\ d=\frac { 32 }{ 75 } \\ e=\frac { 8 }{ 25 } } \\ So,\quad a+b+c+d+e=\frac { 206 }{ 75 } \equiv \frac { p }{ q } \Rightarrow \boxed { 3q-p=19 } .$

The Five Positive Numbers

The answer is 19.

1 solution

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