Question

Show that the numbers $10,11,12$ cannot be the terms of a single $G.P.$ with common ratio not equal to $1$.

Answer 1

Let $T_r=10,T_q=11,T_r=12$
$\therefore {AR}^{p-1}=10,{AR}^{q-1}=11,{AR}^{r-1}=12$
$\therefore R^{p-q}=\frac{10}{11},R^{q-r}=\frac{11}{12}$
or ${\left(\frac{10}{11}\right)}^{1/(p-q)}={\left(\frac{11}{12}\right)}^{1/(q-r)}=R$
or ${\left(\frac{10}{11}\right)}^{q-r}={\left(\frac{11}{12}\right)}^{p-q}$
or ${\left(10\right)}^{q-r}{\left(12\right)}^{p-q}={\left(11\right)}^{p-q+q-r}={\left(11\right)}^{p-r}$
Now $L.H.S.$ of above is clearly even whereas $R.H.S.$ is an odd number which is not possible. Hence $10,11,12$ cannot be the terms of a signle $G.P$.