An algebra problem by Manish Dash

Terms of AP

${ a }_{ p }=a+(p-1)d\\ { a }_{ q }=a+(q-1)d\\ { a }_{ r }=a+(r-1)d\\ { a }_{ s }=a+(s-1)d$

As these terms are in GP so

$\frac { { a }_{ q } }{ { a }_{ p } } =\frac { { a }_{ r } }{ { a }_{ q } } =\frac { { a }_{ s } }{ { a }_{ r } }$

Subtracting 1 from each side we get

$\frac { { a }_{ q }-{ { a }_{ p } } }{ { a }_{ p } } =\frac { { a }_{ r }-{ a }_{ q } }{ { a }_{ q } } =\frac { { a }_{ s }-{ a }_{ r } }{ { a }_{ r } }$

On putting values we get

$\frac { p-q }{ { a }_{ p } } =\frac { q-r }{ { a }_{ q } } =\frac { r-s }{ { a }_{ r } }$

As ${ a }_{ p },{ a }_{ q },{ a }_{ r }$ are in GP so let $a_{p}=A$

So $a_{q}=AR$ and $a_{r}=AR^{2}$

So $\frac { p-q }{ A } =\frac { q-r }{ AR } =\frac { r-s }{ { AR^{ 2 } } }$

So $\frac { p-q }{ A } =\frac { q-r }{ AR }$ and $\frac { q-r }{ AR } =\frac { r-s }{ { AR^{ 2 } } }$

On multiplying these two equations we get

${ \left( \frac { q-r }{ AR } \right) }^{ 2 }=\frac { r-s }{ { AR^{ 2 } } } \frac { p-q }{ A }$

So $\\ (q-r)^{ 2 }=(r-s)(p-q)$

So (p-q), (q-r) and (r-s) are in GP