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QuestionProve that if a,b,ca, b, c are in A.P. and x,y,zx, y, z are in G.P., then xb−c⋅yc−a⋅za−b=1x^{b-c} \cdot y^{c-a} \cdot z^{a-b}=1.

Studdy Solution

STEP 1

Assumptions1. The variables a,b,ca, b, c are in Arithmetic Progression (A.P.) . The variables x,y,zx, y, z are in Geometric Progression (G.P.)
3. We need to prove that xb−c⋅yc−a⋅za−b=1x^{b-c} \cdot y^{c-a} \cdot z^{a-b}=1

STEP 2

In an Arithmetic Progression, the difference between any two consecutive terms is constant. So, we can writeb−a=c−bb - a = c - b

STEP 3

In a Geometric Progression, the ratio of any two consecutive terms is constant. So, we can writeyx=zy\frac{y}{x} = \frac{z}{y}

STEP 4

Now, we will use the property of exponents which states that am⋅an=am+na^{m} \cdot a^{n} = a^{m+n} to rewrite the expression we need to prove.
xb−c⋅yc−a⋅za−b=xb−c⋅yb−2b+c⋅z2b−b−cx^{b-c} \cdot y^{c-a} \cdot z^{a-b} = x^{b-c} \cdot y^{b-2b+c} \cdot z^{2b-b-c}

STEP 5

We can further simplify the above expression by using the property of exponents which states that (am)n=amn(a^{m})^{n} = a^{mn}.
xb−c⋅yb−2b+c⋅z2b−b−c=(xb−c⋅yb−c⋅zb−c)2x^{b-c} \cdot y^{b-2b+c} \cdot z^{2b-b-c} = (x^{b-c} \cdot y^{b-c} \cdot z^{b-c})^{2}

STEP 6

Now, we substitute the values of yy and zz in terms of xx from the geometric progression.
=(xb−c⋅(x⋅yx)b−c⋅(x⋅yx⋅zy)b−c)2= (x^{b-c} \cdot (x \cdot \frac{y}{x})^{b-c} \cdot (x \cdot \frac{y}{x} \cdot \frac{z}{y})^{b-c})^{2}

STEP 7

implify the expression inside the brackets.
=(xb−c⋅xb−c⋅xb−c)2= (x^{b-c} \cdot x^{b-c} \cdot x^{b-c})^{2}

STEP 8

Use the property of exponents which states that am⋅an=am+na^{m} \cdot a^{n} = a^{m+n}.
=(x3(b−c))2= (x^{3(b-c)})^{2}

STEP 9

Use the property of exponents which states that (am)n=amn(a^{m})^{n} = a^{mn}.
=x2⋅3(b−c)= x^{2 \cdot3(b-c)}

STEP 10

Now, substitute the value of b−cb - c from the arithmetic progression.
=x2⋅3⋅0= x^{2 \cdot3 \cdot0}

STEP 11

Any number raised to the power of0 is.
==We have proved that xb−c⋅yc−a⋅za−b=x^{b-c} \cdot y^{c-a} \cdot z^{a-b}=.

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