Question

If $p^{th},q^{th}$ and $r^{th}$ terms of an A.P. and G.P. are both a, b and c respectively, show that $a^{b-c}{b}^{c-a}{c}^{a-b}=1$.

Answer 1

Let A be the first term and D be the common difference of the A.P. Therefore,

$a_p=A+(p-1)D=a$ ..........(i)

$a_q=A+(q-1)D=b$ ...........(ii)

$a_r=A+(r-1)D=c$ ...........(iii)

Also, suppose A' be the first and R be the common ratio of the G.P. Therefore,

$a_p=A^{\prime }{R}^{p-1}=a$ ..........(iv)

$a_q=A^{\prime }{R}^{q-1}=b$ ...........(v)

$a_r=A^{\prime }{R}^{r-1}=c$ ...........(vi)

Now,

Subtracting (ii) from (i), we get.

$A+(p-1)D-A-\left)\right(q-1)D=a-b$

$\Rightarrow (p-q)D=a-b$ ........(vii)

Subtracting (iii) from (ii), we get

$A+(q-1)D-A-(r-1)D=b-c$

$\Rightarrow (q-r)D=b-c$ .........(ix)

$\therefore a^{b-c}{b}^{c-a}{c}^{a-b}$

$[A^{\prime }{R}^{(p-1)}{]}^{(q-r)D}\times [A^{\prime }{R}^{(q-1)}{]}^{(r-p)D}\times [A^{\prime }{R}^{r-1}{]}^{(p-q)D}$

[Using (iv), (v), (vi), (vii), (viii), (ix)]

$=A^{\prime (q-r)D}{R}^{(p-1)(q-r)D}\times A^{\prime (r-p)D}{R}^{(q-1)(r-p)D}\times A^{\prime }(p-q)D{R}^{(r-1)(p-q)D}$

$=A^{\prime \left[\right(q-r)D+(r-p)D+(p+q\left)D\right]\times R^{\left[\right(p-1\left)\right(q-r)D+(q-1\left)\right(r-p)D+(r-1\left)\right(p-q\left)D\right]}}$

$=A^{\prime [q-r+r-p+p-q]D}\times R^{[pq-pr-q+r+qr-pq-r+p+pr-qr-p+q]D}$

$=(A^{\prime }{)}^0\times R^0$

$=1\times 1=1$