Sequence

Algebra Level 3

Given that x 1 , x 2 , … , x 16 x_1, x_2, \ldots,x_{16} are positive real numbers such that x 1 x 2 = x 2 x 3 = ⋯ = x 15 x 16 \dfrac{x_1}{x_2} = \dfrac{x_2}{x_3} = \cdots = \dfrac{x_{15}} {x_{16}} .

If x 1 + x 2 + x 3 + x 4 = 20 x_1 + x_2 + x_3 + x_4 = 20 and x 5 + x 6 + x 7 + x 8 = 320 x_5 + x_6 + x_7 +x_8 = 320 , find x 13 + x 14 + x 15 + x 16 x_{13} + x_{14} + x_{15} + x_{16} .

31290 12390 98120 81920

Amish Garg by Amish Garg , 20, India

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1 solution

Hassan Abdulla
Aug 22, 2017

x 1 x 2 = x 2 x 3 = x 3 x 4 = . . . . . . . . . . = x 15 x 16 \frac { { x }_{ 1 } }{ { x }_{ 2 } } =\frac { { x }_{ 2 } }{ { x }_{ 3 } } =\frac { { x }_{ 3 } }{ { x }_{ 4 } } =..........=\frac { { x }_{ 15 } }{ { x }_{ 16 } }

x 2 x 1 = x 3 x 2 = x 4 x 3 = . . . . . . . . . . = x 16 x 15 = α \frac { { x }_{ 2 } }{ { x }_{ 1 } } =\frac { { x }_{ 3 } }{ { x }_{ 2 } } =\frac { { x }_{ 4 } }{ { x }_{ 3 } } =..........=\frac { { x }_{ 16 } }{ { x }_{ 15 } } =\alpha

x 2 x 1 = α \frac { { x }_{ 2 } }{ { x }_{ 1 } } =\alpha

x 3 x 1 = x 3 x 2 ⋅ x 2 x 1 = α ⋅ α = α 2 \frac { { x }_{ 3 } }{ { x }_{ 1 } } =\frac { { x }_{ 3 } }{ { x }_{ 2 } } \cdot \frac { { x }_{ 2 } }{ { x }_{ 1 } } =\alpha \cdot \alpha ={ \alpha }^{ 2 }

x 4 x 1 = x 4 x 3 ⋅ x 3 x 2 ⋅ x 2 x 1 = α 3 \frac { { x }_{ 4 } }{ { x }_{ 1 } } =\frac { { x }_{ 4 } }{ { x }_{ 3 } } \cdot \frac { { x }_{ 3 } }{ { x }_{ 2 } } \cdot \frac { { x }_{ 2 } }{ { x }_{ 1 } } ={ \alpha }^{ 3 }

⋮ \vdots

x k x 1 = α k − 1 ⟹ x k = α k − 1 x 1 → ( 1 ) \frac { { x }_{ k } }{ { x }_{ 1 } } ={ \alpha }^{ k-1 }\Longrightarrow { x }_{ k }={ \alpha }^{ k-1 }{ x }_{ 1 }\rightarrow \left( 1 \right)

x 1 + x 2 + x 3 + x 4 = 20 { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }+{ x }_{ 4 }=20

x 1 + α x 1 + α 2 x 1 + α 3 x 1 = 20 { x }_{ 1 }+{ \alpha x }_{ 1 }+{ { \alpha }^{ 2 }x }_{ 1 }+{ { \alpha }^{ 3 }x }_{ 1 }=20 substitute using (1)

x 1 ( 1 + α 1 + α 2 + α 3 ) = 20 ⟹ x 1 = 20 ( 1 + α 1 + α 2 + α 3 ) → ( 2 ) { x }_{ 1 }\left( 1+{ \alpha }_{ 1 }+{ { \alpha }^{ 2 } }+{ { \alpha }^{ 3 } } \right) =20\Longrightarrow { x }_{ 1 }=\frac { 20 }{ \left( 1+{ \alpha }_{ 1 }+{ { \alpha }^{ 2 } }+{ { \alpha }^{ 3 } } \right) } \rightarrow \left( 2 \right)

similarly

x 5 + x 6 + x 7 + x 8 = 320 { x }_{ 5 }+{ x }_{ 6 }+{ x }_{ 7 }+{ x }_{ 8 }=320

α 4 x 1 + α 5 x 1 + α 6 x 1 + α 7 x 1 = 320 { { \alpha }^{ 4 }x }_{ 1 }+{ { \alpha }^{ 5 }x }_{ 1 }+{ { \alpha }^{ 6 }x }_{ 1 }+{ { \alpha }^{ 7 }x }_{ 1 }=320

α 4 x 1 ( 1 + α + α 2 + α 3 ) = 320 { { \alpha }^{ 4 }x }_{ 1 }\left( 1+{ \alpha }+{ { \alpha }^{ 2 } }+{ { \alpha }^{ 3 } } \right) =320

α 4 ⋅ 20 ( 1 + α + α 2 + α 3 ) ⋅ ( 1 + α 1 + α 2 + α 3 ) = 320 ⟹ α = 2 { { \alpha }^{ 4 }\cdot \frac { 20 }{ \left( 1+\alpha +{ { \alpha }^{ 2 } }+{ { \alpha }^{ 3 } } \right) } \cdot }\left( 1+{ \alpha }_{ 1 }+{ { \alpha }^{ 2 } }+{ { \alpha }^{ 3 } } \right) =320\Longrightarrow \alpha =2 substitute using (2)

x 1 = 20 ( 1 + 2 + 2 2 + 2 3 ) = 4 3 { x }_{ 1 }=\frac { 20 }{ \left( 1+2+{ 2^{ 2 } }+{ 2^{ 3 } } \right) } =\frac { 4 }{ 3 } substitute α { \alpha } in (2)

Now : x 13 + x 14 + x 15 + x 16 = α 12 x 1 ( 1 + α + α 2 + α 3 ) = 81920 { x }_{ 13 }+{ x }_{ 14 }+{ x }_{ 15 }+{ x }_{ 16 }={ { \alpha }^{ 12 }x }_{ 1 }\left( 1+{ \alpha }+{ { \alpha }^{ 2 } }+{ { \alpha }^{ 3 } } \right) =81920

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