Question

# If the product of three consecutive terms is G.P. is $$216$$ and sum of their products is pairs is $$156$$, find them

Solution

## Let $$\dfrac { a }{ r } ,a,ar$$ be the first three terms of an G.P. It is given that the product of the terms is $$216$$ that is:$$\dfrac { a }{ r } \times a\times ar=216\\ \Rightarrow { a }^{ 3 }=216\\ \Rightarrow { a }^{ 3 }=6^{ 3 }\\ \Rightarrow { a }=6$$It is also given that the sum of their products is $$156$$ that is:$$\left( \dfrac { a }{ r } \times a \right) +\left( a\times ar \right) +\left( ar\times \dfrac { a }{ r } \right) =156\\ \Rightarrow \dfrac { a^{ 2 } }{ r } +a^{ 2 }r+a^{ 2 }=156\\ \Rightarrow a^{ 2 }\left( \dfrac { 1 }{ r } +r+1 \right) =156\\ \Rightarrow a^{ 2 }\left( \dfrac { 1+{ r }^{ 2 }+r }{ r } \right) =156\\ \Rightarrow a^{ 2 }\left( \dfrac { { r }^{ 2 }+r+1 }{ r } \right) =156$$Substituting $$a=6$$, we get,$$6^{ 2 }\left( \dfrac { { r }^{ 2 }+r+1 }{ r } \right) =156\\ \Rightarrow 36\left( \dfrac { { r }^{ 2 }+r+1 }{ r } \right) =156\\ \Rightarrow \dfrac { { r }^{ 2 }+r+1 }{ r } =\dfrac { 156 }{ 36 } \\ \Rightarrow \dfrac { { r }^{ 2 }+r+1 }{ r } =\dfrac { 13 }{ 3 }$$  $$\Rightarrow 3({ r }^{ 2 }+r+1)=13r\\ \Rightarrow 3{ r }^{ 2 }+3r+3=13r\\ \Rightarrow 3{ r }^{ 2 }+3r-13r+3=0\\ \Rightarrow 3{ r }^{ 2 }-10r+3=0\\ \Rightarrow 3{ r }^{ 2 }-9r-r+3=0$$$$\Rightarrow 3r(r-3)-1(r-3)=0\\ \Rightarrow (3r-1)(r-3)=0\\ \Rightarrow 3r-1=0,\quad r-3=0\\ \Rightarrow 3r=1,\quad r=3\\ \Rightarrow r=\dfrac { 1 }{ 3 } ,\quad r=3$$ Now, if $$a=6$$ and $$r=\dfrac { 1 }{ 3 }$$ then the first three terms of the G.P are: $$\dfrac { a }{ r } =\dfrac { 6 }{ \dfrac { 1 }{ 3 } } =6\times3=18$$$$a=6$$ and $$ar=6\times \dfrac { 1 }{ 3 } =2$$ And if $$a=6$$ and $$r=3$$ then the first three terms of the G.P are:$$\dfrac { a }{ r } =\dfrac { 6 }{ 3 } =2$$$$a=6$$ and $$ar=6\times 3 =18$$  Hence, the three terms are $$18,6,2$$ or $$2,6,18$$Quantitative Aptitude

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