$b = ar, c = ar^2$  અને  $= ar^3$  લેતાં, પછી 

$\frac{{\text{1}}}{{{{\text{a}}^{\text{3}}} + {{\text{b}}^{\text{3}}}}} = \frac{{\text{1}}}{{{{\text{a}}^{\text{3}}}\left( {{\text{1}} + {{\text{r}}^{\text{3}}}} \right)}}{\text{,}}\,\,\frac{{\text{1}}}{{{{\text{b}}^{\text{3}}} + {{\text{c}}^{\text{3}}}}} = \frac{{\text{1}}}{{{{\text{a}}^{\text{3}}}{{\text{r}}^{\text{3}}}\left( {{\text{1}} + {{\text{r}}^{\text{3}}}} \right)}}$ અને $\frac{{\text{1}}}{{{{\text{c}}^{\text{3}}} + {{\text{d}}^{\text{3}}}}} = \frac{{\text{1}}}{{{{\text{a}}^{\text{3}}}{{\text{r}}^{\text{6}}}\left( {{\text{1}} + {{\text{r}}^{\text{3}}}} \right)}}$

તેથી, ${\left( {{{\text{a}}^{\text{3}}} + {{\text{b}}^{\text{3}}}} \right)^{{\text{ – 1}}}}, {\text{ }}{\left( {{{\text{b}}^{\text{3}}} + {{\text{c}}^{\text{3}}}} \right)^{{\text{ – 1}}}}$ અને $ {\left( {{{\text{c}}^{\text{3}}} + {{\text{d}}^{\text{3}}}} \right)^{ – {\text{1}}}}\,$

સમગુણોત્તર શ્રેણીમાં સમાન તફાવત $\frac{{\text{1}}}{{{{\text{r}}^{\text{3}}}}}$ સાથે છે.