(A&G)P=AP and GP

Algebra Level 2

Given that $a,b$ and $c$ are in an arithmetic progression ;
$x,y$ and $z$ are in a geometric progression .
Find the value of

$\Large x^{b-c} \times y^{c-a} \times z^{a-b} .$

The answer is 1.

2 solutions

Rohit Roychoudhury
May 22, 2016

$a, b, c$ are in AP, then $a - b = b - c$
$x, y, z$ are in GP, then $y^2=xz$

Therefore :
$\begin{aligned} x^{b-c} \ y^{c-a} \ z^{a-b} & = x^{b-c} \ y^{c-a} \ z^{b-c} \\ & =(xz)^{b-c} \ y^{c-a} \\ & =y^{2(b-c)} \ y^{c-a} \\ & = y^{2(b-c)+(c-a)} \\ & = y^{2b-a-c} \\ & = y^0 = 1 \end{aligned}$

Thanks for sharing this problem and solution, Rohit. I have edited the Latex in the solution so it is easier for others to read and understand.

Pranshu Gaba - 5 years ago

Thanx pranshu for the editing

Rohit Roychoudhury - 5 years ago

we can also solve it by taking the common difference d and the common ratio as k and substituting them in the equation properly.

Piyush Kumar Behera - 5 years ago

Hung Woei Neoh
May 22, 2016

Given an AP $a,b,c,\ldots$

We let the common difference be $d$

We can say that $b=a+d,\;c=a+2d$

Given a GP $x,y,z,\ldots$

Similarly, we let the common ratio be $r$

We can say that $y=xr,\;z=xr^2$

Now, we want to evaluate this:

$\large x^{b-c} \times y^{c-a} \times z^{a-b}\\ \large =x^{(a+d) - (a+2d)} \times (xr)^{(a+2d)-a} \times (xr^2)^{a-(a+d)}\\ \large =x^{-d} \times (xr)^{2d} \times (xr^2)^{-d}\\ \large = \dfrac{1}{x^d} \times x^{2d}r^{2d} \times \dfrac{1}{x^dr^{2d}}\\ \large = \dfrac{x^{2d}r^{2d}}{x^{2d}r^{2d}}\\ \large = \boxed{1}$

(A&G)P=AP and GP

The answer is 1.

2 solutions

0 pending reports