Question

Show that the following series is in Geometric progression (GP)
$\left( {{a^2} + {b^2} + {c^2}} \right),\left( {ab + bc + cd} \right),\left( {{b^2} + {c^2} + {d^2}} \right)$

Answer 1

Hint:Start the solution by defining geometric progression. Make use of suitable substitutions in the given series. After this step as the definition states, find the ratios of the adjacent term. This ratio should be the same for the entire given series and that completes the proof.

Complete step by step answer:
to show that the given series is in GP we will first try to understand the definition of GP.
GP is a sequence in which each term except the first term is obtained by multiplying the previous term by a non-zero constant called common ratio. We will start the solution by substituting $b = ar,c = a{r^2},d = a{r^3}$. So the first term becomes,
$\left( {{a^2} + {b^2} + {c^2}} \right) = \left( {{a^2} + {a^2}{r^2} + {a^2}{r^4}} \right) - - - \left( 1 \right)$
$\Rightarrow \left( {ab + bc + cd} \right) = \left( {{a^2}r + {a^2}{r^3} + {a^2}{r^5}} \right) - - - \left( 2 \right)$
\[\Rightarrow \left( {{b^2} + {c^2} + {d^2}} \right) = \left( {{a^2}{r^2} + {a^2}{r^4} + {a^2}{r^6}} \right) - - - \left( 3 \right)\]
Now since the initial term should be multiple of second, we start with first two terms of the series therefore,
$\dfrac{{\left( {ab + bc + cd} \right)}}{{\left( {{a^2} + {b^2} + {c^2}} \right)}} = \dfrac{{\left( {{a^2}r + {a^2}{r^3} + {a^2}{r^5}} \right)}}{{\left( {{a^2} + {a^2}{r^2} + {a^2}{r^4}} \right)}}$
$\Rightarrow\dfrac{{\left( {ab + bc + cd} \right)}}{{\left( {{a^2} + {b^2} + {c^2}} \right)}}= \dfrac{{{a^2}r\left( {1 + {r^2} + {r^4}} \right)}}{{{a^2}\left( {1 + {r^2} + {r^4}} \right)}} \\
\Rightarrow\dfrac{{\left( {ab + bc + cd} \right)}}{{\left( {{a^2} + {b^2} + {c^2}} \right)}}= r - - - \left( 4 \right) \\ $
Also,
$\dfrac{{\left( {{b^2} + {c^2} + {d^2}} \right)}}{{\left( {ab + bc + cd} \right)}} = \dfrac{{\left( {{a^2}{r^2} + {a^2}{r^4} + {a^2}{r^6}} \right)}}{{\left( {{a^2}r + {a^2}{r^3} + {a^2}{r^5}} \right)}}$
\[\Rightarrow\dfrac{{\left( {{b^2} + {c^2} + {d^2}} \right)}}{{\left( {ab + bc + cd} \right)}}= \dfrac{{{a^2}{r^2}\left( {1 + {r^2} + {r^4}} \right)}}{{{a^2}r\left( {1 + {r^2} + {r^4}} \right)}} \\
\therefore\dfrac{{\left( {{b^2} + {c^2} + {d^2}} \right)}}{{\left( {ab + bc + cd} \right)}}= r - - - \left( 5 \right) \\ \]
Hence from (4) and (5) we can see that the ratios are the same.Thus the given series is a GP.

Note:If you have been given to check if the series is geometric, always strive to find the common ratio between the two adjacent terms. If the ratios are the same then we can easily conclude that the series is geometric. If we are unable to find a common ratio then the series cannot be a geometric progression.