# If pth, qth, and rth terms of an A.P. and G.P.

Question:

If pth, qth, and rth terms of an A.P. and G.P. are both a, b and c respectively, show that

ab–c . bc – a. ca – b = 1

Solution:

Let the first term of AP be $m$ and common difference as $d$

Let the GP first term as I and common ratio as s

The $\mathrm{n}^{\text {th }}$ term of an AP is given as $\mathrm{t}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}$ where $\mathrm{a}$ is the first term and $\mathrm{d}$ is the common difference

The $n^{\text {th }}$ term of $a$ GP is given by $t_{n}=a r^{n-1}$ where $a$ is the first term and $r$ is the common ratio

The $p^{\text {th }}$ term $\left(t_{p}\right)$ of both AP and GP is a

For AP

$\Rightarrow \mathrm{t}_{\mathrm{p}}=\mathrm{m}+(\mathrm{p}-1) \mathrm{d}$

$\Rightarrow \mathrm{a}=\mathrm{m}+(\mathrm{p}-1) \mathrm{d} \ldots . .1$

For GP

$\Rightarrow \mathrm{t}_{\mathrm{p}}=\mathrm{Is}^{\mathrm{p}-1}$

$\Rightarrow \mathrm{a}=\mathrm{Is}^{\mathrm{p}-1} \ldots .2$

The $q^{\text {th }}$ term $\left(t_{q}\right)$ of both AP and GP is b For AP

$\Rightarrow \mathrm{t}_{\mathrm{q}}=\mathrm{m}+(\mathrm{q}-1) \mathrm{d}$

$\Rightarrow \mathrm{b}=\mathrm{m}+(\mathrm{q}-1) \mathrm{d} \ldots .3$

For GP

$\Rightarrow \mathrm{t}_{\mathrm{q}}=1 \mathrm{~s}^{\mathrm{q}-1}$

$\Rightarrow \mathrm{b}=\mathrm{Is}^{\mathrm{q}-1} \ldots . .4$

The $r^{\text {th }}$ term $\left(t_{r}\right)$ of both AP and GP is c For AP

$\Rightarrow \mathrm{t}_{\mathrm{r}}=\mathrm{m}+(\mathrm{r}-1) \mathrm{d}$

$\Rightarrow \mathrm{c}=\mathrm{m}+(\mathrm{r}-1) \mathrm{d} \ldots .5$

For GP

$\Rightarrow \mathrm{t}_{\mathrm{r}}=\mathrm{s}^{\mathrm{r}-1}$

$\Rightarrow \mathrm{c}=\mathrm{Is}^{\mathrm{r}-1} \ldots \ldots 6$

Let us find $b-c, c-a$ and $a-b$

Using 3 and 5

$\Rightarrow \mathrm{b}-\mathrm{c}=(\mathrm{q}-\mathrm{r}) \mathrm{d} \ldots$ (i)

Using 5 and 1

$\Rightarrow c-a=(r-p) d \ldots$ (ii)

Using 1 and 3

$\Rightarrow a-b=(p-q) d \ldots$ (iii)

We have to prove that $a^{b-c} \cdot b^{c-a} \cdot c^{a-b}=1$

LHS $\left.=a^{b-c} \cdot b^{c-a} \cdot c^{a-b}\right)$

$\Rightarrow c-a=(r-p) d \ldots$ (ii)

Using 1 and 3

$\Rightarrow \mathrm{a}-\mathrm{b}=(\mathrm{p}-\mathrm{q}) \mathrm{d} \ldots$ (iii)

We have to prove that $a^{b-c} \cdot b^{c-a} \cdot c^{a-b}=1$

$\left.\mathrm{LHS}=\mathrm{a}^{\mathrm{b}-\mathrm{c}} \cdot \mathrm{b}^{\mathrm{c}-\mathrm{a}} \cdot \mathrm{c}^{\mathrm{a}-\mathrm{b}}\right)$

$=\left(\frac{1 s^{p}}{s}\right)^{b-c} \cdot\left(\frac{1 s^{q}}{s}\right)^{c-a} \cdot\left(\frac{1 s^{r}}{s}\right)^{a-b}$

$=\frac{1^{\mathrm{b}-\mathrm{c}+\mathrm{c}-\mathrm{a}+\mathrm{a}-\mathrm{b}}}{\mathrm{s}^{\mathrm{b}-\mathrm{c}+\mathrm{c}-\mathrm{a}+\mathrm{a}-\mathrm{b}}} \cdot \mathrm{S}^{\mathrm{p}(\mathrm{b}-\mathrm{c})} \cdot \mathrm{S}^{\mathrm{q}(\mathrm{c}-\mathrm{a})} \cdot \mathrm{S}^{\mathrm{r}(\mathrm{a}-\mathrm{b})}$

Substituting values of $a-b, c-a$ and $b-c$ from (iii), (ii) and (i)

$=S^{p(q-r) d} \cdot S^{q(r-p) d} \cdot S^{r(p-q) d}$

$=S^{p q d-p r d} \cdot S^{q r d-p q d} \cdot S^{p r d-q r d}$

$=S^{p q d-p r d+q r d-p q d+p r d-q r d}$

$=s^{0}=1$

$\Rightarrow \mathrm{LHS}=\mathrm{RHS}$

Hence proved

Objective Type Questions