Question

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

Answer 1

Step $1$. Using given data.

Let assume first term is $a$ and common difference is $d$.

$p^{th}\text{term of A.P.}=a_p=a+(p-1)d$

$q^{th}\text{term of A.P.}=a_q=a+(q-1)d$

$r^{th}\text{term of A.P.}=a_r=a+(r-1)d$

$s^{th}\text{term of A.P.}=a_s=a+(s-1)d$

It is given that $a_p,a_q,a_r$ & $a_s$ are in $G.P.$

So, their common ratio is same

$\frac{a_q}{a_p}=\frac{a_r}{a_q}=\frac{a_s}{a_r}$ $....\left(i\right)$

Step $2$. Simplifies.

$\frac{a_q}{a_p}=\frac{a_r}{a_q}=\frac{a_s}{a_r}$ from $\left(i\right)$

Now solving,

$\frac{a_q}{a_p}=\frac{a_r}{a_q}$

Subtracting $1$ both sides,

$\frac{a_q}{a_p}-1=\frac{a_r}{a_q}-1$

$\Rightarrow \frac{(a_q-a_p)}{a_p}=\frac{(a_r-a_q)}{a_q}$

$\Rightarrow \frac{(a_q-a_p)}{(a_r-a_q)}=\frac{a_p}{a_q}$$....\left(ii\right)$

$\frac{a_q}{a_p}=\frac{a_r}{a_q}=\frac{a_s}{a_r}....\left(i\right)$

$\frac{(a_q-a_p)}{(a_r-a_q)}=\frac{a_p}{a_q}....\left(ii\right)$

Again

$\frac{a_r}{a_q}-1=\frac{a_s}{a_r}-1$

$\Rightarrow \frac{(a_r-a_q)}{a_q}=\frac{(a_s-a_r)}{a_r}$

$\Rightarrow \frac{(a_s-a_r)}{(a_r-a_q)}=\frac{a_r}{a_q}$

$\Rightarrow \frac{(a_r-a_s)}{(a_q-a_r)}=\frac{a_r}{a_q}$ $...\left(iii\right)$

$\frac{a_q}{a_p}=\frac{a_r}{a_q}=\frac{a_s}{a_r}...\left(i\right)$

$\frac{(a_q-a_p)}{(a_r-a_q)}=\frac{a_p}{a_q}....\left(ii\right)$

$\frac{(a_r-a_s)}{(a_q-a_r)}=\frac{a_r}{a_q}...\left(iii\right)$

Step $3$. Solve tp prove $(p-q),(q-r),(r-s)$ are in G.P

Substitute values of $a_p,a_q$ & $a_r$ in $\left(ii\right)$

$\Rightarrow \frac{a+(q-1)d-[a+(p-1\left)d\right]}{a+(r-1)d-[a+(q-1\left)d\right]}=\frac{a_p}{a_q}$

$\Rightarrow \frac{a-a+(q-1)d-(p-1)d]}{a-a+(r-1)d-(q-1)d]}=\frac{a_p}{a_q}$

$\Rightarrow \frac{d[q-1)-(p-1)]}{d\left[\right(r-1)-(q-1\left)\right]}=\frac{a_p}{a_q}$

$\Rightarrow \frac{q-1-p+1}{r-1-q+1}=\frac{a_p}{a_q}$

$\Rightarrow \frac{q-p}{r-q}=\frac{a_p}{a_q}$

$\frac{a_q}{a_p}=\frac{a_r}{a_q}=\frac{a_s}{a_r}...\left(i\right)$

$\frac{(a_r-a_s)}{(a_q-a_r)}=\frac{a_r}{a_q}....\left(iii\right)$

$\frac{q-p}{r-q}=\frac{a_p}{a_q}$

$\Rightarrow \frac{r-q}{q-p}=\frac{a_q}{a_p}$

$\Rightarrow \frac{-(q-r)}{-(p-q)}=\frac{a_q}{a_p}$

$\Rightarrow \frac{(q-r)}{(p-q)}=\frac{a_q}{a_p}$ $...\left(iv\right)$

Again putting values $a_q,a_r$ & $a_s$ in $\left(iii\right)$

$\frac{a+(r-1)d-[a+(s-1\left)d\right]}{a+(q-1)d-[a+(r-1\left)d\right]}=\frac{a_r}{a_q}$

$\Rightarrow \frac{a-a+(r-1)d-(s-1)d}{a-a+(q-1)d-(r-1)d}=\frac{a_r}{a_q}$

$\Rightarrow \frac{d\left[\right(r-1)-(s-1\left)\right]}{d\left[\right(q-1)-(r-1\left)\right]}=\frac{a_r}{a_q}$


$\Rightarrow \frac{r-1-s+1}{q-1-r+1}=\frac{a_r}{a_q}$

$\Rightarrow \frac{r-s}{q-r}=\frac{a_r}{a_q}$$...\left(v\right)$ $\left\{\text{from}\right(i\left)\right\}$


$\text{From}\left(iv\right)\text{and}\left(v\right)$

$\frac{(q-r)}{(p-q)}=\frac{r-s}{q-r}$

$\because \text{common ratio for}(p-q),(q-r),(r-s)\text{is same i.e,}\frac{(q-r)}{(p-q)}=\frac{r-s}{q-r}$

$\therefore (p-q),(q-r),(r-s)$ are in G.P.