# Show that The following series is in GP ( Geometric Progression ) ?

## $\left({a}^{2} + {b}^{2} + {c}^{2}\right) , \left(a b + b c + c d\right) , \left({b}^{2} + {c}^{2} + {d}^{2}\right)$ Prove It is in Geometric Progression

Aug 8, 2018

#### Explanation:

I think the Problem is to prove :

If $a , b , c , d$ are in GP, then, show that,

$\left({a}^{2} + {b}^{2} + {c}^{2}\right) , \left(a b + b c + c d\right) , \left({b}^{2} + {c}^{2} + {d}^{2}\right)$ are also in GP.

It suffices to show that,

${\left(a b + b c + c d\right)}^{2} = \left({a}^{2} + {b}^{2} + {c}^{2}\right) \left({b}^{2} + {c}^{2} + {d}^{2}\right) \ldots \ldots \ldots . \left(\star\right)$.

Given that, $a , b , c , d$ are in GP.

$\therefore \frac{b}{a} = \frac{c}{b} = \frac{d}{c} = r , \text{ say}$.

$\therefore b = a r , c = b r = \left(a r\right) r = a {r}^{2} , \mathmr{and} , d = c r = a {r}^{3.} . . \left(\star '\right)$.

Utilising $\left(\star '\right)$ in the R.H.S. of $\left(\star\right)$, we have,

$\text{The R.H.S. of } \left(\star\right) = \left({a}^{2} + {b}^{2} + {c}^{2}\right) \left({b}^{2} + {c}^{2} + {d}^{2}\right)$,

$= \left({a}^{2} + {a}^{2} {r}^{2} + {a}^{2} {r}^{4}\right) \left({a}^{2} {r}^{2} + {a}^{2} {r}^{4} + {a}^{2} {r}^{6}\right)$,

$= {a}^{2} \left(1 + {r}^{2} + {r}^{4}\right) \left\{{a}^{2} {r}^{2} \left(1 + {r}^{2} + {r}^{4}\right)\right\}$,

$= {a}^{4} {r}^{2} {\left(1 + {r}^{2} + {r}^{4}\right)}^{2}$,

$= {\left\{{a}^{2} r \left(1 + {r}^{2} + {r}^{4}\right)\right\}}^{2}$,

$= {\left({a}^{2} r + {a}^{2} {r}^{3} + {a}^{2} {r}^{5}\right)}^{2}$,

$= {\left(a \cdot a r + a r \cdot a {r}^{2} + a {r}^{2} \cdot a {r}^{3}\right)}^{2}$,

$= {\left(a b + b c + c d\right)}^{2}$,

$\text{=The L.H.S. of } \left(\star\right)$.

Hence, the Proof.