Question

5. If p. q and r are in A.P. and x, y, z are in G.P. then x^q-r × y^r-p × z^p-q is equal to
(a) 0
(b)-1
(c) 1
(d) none of these​

Answer 1

Answer:

1

Step-by-step explanation:

If p, q are r are in AP, 2q = p + r

Therefore, q - r = p - q

If x, y and z are in GP, y² = xz

Now the question looks like:

= x^q-r × y^r-p × z^p-q

= x^p-q × y^r-p × z^p-q

= (xz)^p-q × y^r-p

= (y²)^p-q × y^r-p

= y^2p-2q × y^r-p

= y^(2p - 2q + r - p)

= y^(p + r - 2q)

= y^(0)

= 1

Line 2: x^q-r is changed with x^p-q, as q-r = p-q.

Line 3: since x and z have same power, they can be multiplied.

Line 4: xz = y², condition for GP.

Line 8: p + r = 2q, so p + r - 2q = 0.

Answer 2

Given that, The p , q & r are in A.P. [ Airthmetic Progression ] , and x, y & z are in G.P [ Geometric Progression ] .

Exigency To Find : The Value of $\bf x^{q - r} \times y^{r-p} \times z^{p -q}$ ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀Given that ,

⠀⠀⠀⠀⠀⠀▪︎⠀The p , q & r are in A.P. [ Airthmetic Progression ].

Therefore,

$\qquad \dashrightarrow \sf 2q \:=\: p + r$

$\qquad \dashrightarrow \sf q \:=\: \dfrac{p + r}{2}$

$\qquad \dashrightarrow \bf q \:=\: \dfrac{p + r}{2} \qquad \:\:\bigg\lgroup \sf{\:Equation \: 1 \:}\bigg\rgroup$

⠀⠀⠀⠀⠀⠀⠀⠀AND ,

⠀⠀⠀⠀⠀⠀▪︎⠀The x , y & z are in G.P. [ Geometric Progression ].

Therefore,

• Condition for an G.P [ Geometric Progression] if x , y & z are in G.P. [ Geometric Progression ].

$\qquad \dashrightarrow \sf y^2 \:=\: x z$

$\qquad \dashrightarrow \sf y^2 \:=\: x z$

$\qquad \dashrightarrow \bf y^2 \:=\: x z \qquad \:\:\bigg\lgroup \sf{\:Equation \: 2 \:}\bigg\rgroup$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ Finding the value of $\bf x^{q - r} \times y^{r-p} \times z^{p -q}$ :

$\qquad \dashrightarrow \sf x^{q - r} \times y^{r-p} \times z^{p -q}$

As We know that ,

$\qquad \dashrightarrow \bf q \:=\: \dfrac{p + r}{2} \qquad \:\:\bigg\lgroup \sf{\:Equation \: 1 \:}\bigg\rgroup$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: value\: of \: q \: as \: Equation \: 1 \::}}$

$\qquad \dashrightarrow \sf x^{q - r} \times y^{r-p} \times z^{p -q}$

$\qquad \dashrightarrow \sf x^{\Big\{ \dfrac{ p + r}{2} - r\Big\} } \times y^{r-p} \times z^{\Big\{p - \dfrac{ p + r}{2} \Big\}}$

$\qquad \dashrightarrow \sf x^{\Big\{ \dfrac{ p - r}{2} \Big\} } \times y^{r-p} \times z^{\Big\{ \dfrac{ p - r}{2} \Big\}}$

As , We know that ,

$\qquad \dashrightarrow \bf y^2 \:=\: x z \qquad \:\:\bigg\lgroup \sf{\:Equation \: 2 \:}\bigg\rgroup$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: value\: of \: y \: as \: Equation \: 2 \::}}$

$\qquad \dashrightarrow \sf x^{\Big\{ \dfrac{ p - r}{2} \Big\} } \times y^{r-p} \times z^{\Big\{ \dfrac{ p - r}{2} \Big\}}$

$\qquad \dashrightarrow \sf x^{\Big\{ \dfrac{ p - r}{2} \Big\} } \times (xz)^{\Big\{ \dfrac{ r - p }{2} \Big\}} \times z^{\Big\{ \dfrac{ p - r}{2} \Big\}}$

$\qquad \dashrightarrow \sf x^{\Big\{ \dfrac{ p - r}{2} \Big\} } \times x^{\Big\{ \dfrac{ r - p }{2} \Big\}} \times z^{\Big\{ \dfrac{ r - p }{2} \Big\}} \times z^{\Big\{ \dfrac{ p - r}{2} \Big\}}$

$\qquad \dashrightarrow \sf x^0 + z^0$

$\qquad \dashrightarrow \sf 1$

$\qquad \dashrightarrow \underline {\boxed {\pmb{\frak{\pink{ \: x^{q - r} \times y^{r-p} \times z^{p -q}\:\:=\:1\:}}}}}\:\bigstar \:$

$\qquad \therefore \:\underline {\sf Hence,\: The \: Value \:of \: x^{q - r} \times y^{r-p} \times z^{p -q}\:is \: \pmb{\bf 1 }\:.}$