Question

If $p^{th},q^{th},r^{th}$ and $s^{th}$, terms of an A.P are in G.P, then the show that $(p-q),(q-r),(r-s)$ are also in G.P ?

Answer 1

$\because a_p=a+(p-1)d;a_q=a+(q-1)d;a_r=a+(r-1)d;a_s=a+(s-1)d;A/Q{p}^{th},q^{th},r^{th},s^{th}termareinG.P\therefore \frac{a_q}{a_p}=\frac{a_r}{a_q}=\frac{a_s}{a_r}for\frac{a_q}{a_p}=\frac{a_r}{a_q}\Rightarrow \frac{a_q}{a_p}-1=\frac{a_r}{a_q}-1\Rightarrow \frac{a_q}{a_p}=\frac{a_r-a_q}{a_q-a_p}puttingthevalueoftermweget\Rightarrow \frac{a_q}{a_p}=\frac{(a+(r-1\left)d\right)-(a+(q-1\left)d\right)}{(a+(q-1\left)d\right)-(a+(p-1\left)d\right)}=\frac{q-r}{p-q}⟶\left(1\right)similarlyfor\frac{a_r}{a_q}=\frac{a_s}{a_r}\Rightarrow \frac{a_r}{a_q}=\frac{r-s}{q-r}⟶\left(2\right)from\left(1\right)and\left(2\right)wecometoknow(p-q),(q-r)and(r-s)areinG.P$