Question

if a,b,c are in G.p and a^1/x= b^1/y= c^1/z then prove that x,y,z are in a.p.​

Answer 1

EXPLANATION.

a, b, c are in G.P.

$\sf \implies a^{\dfrac{1}{x} } = b^{\dfrac{1}{y} } = c^{\dfrac{1}{z} }$

If a, b, c are in G.P.

As we know that,

Conditions of G.P.

⇒ b² = ac.

Take log on both sides, we get.

⇒ ㏒(b)² = ㏒(ac).

⇒ 2㏒(b) = ㏒(a) + ㏒(c). ⇒ (1).

Same as equation,

$\sf \implies a^{\dfrac{1}{x} } = b^{\dfrac{1}{y} } = c^{\dfrac{1}{z} }$

Taking log on this equation, we get.

⇒ ㏒(a)/x = ㏒(b)/y = ㏒(c)/z. ⇒ (2).

From equation (1) and (2) we get,

From equation (1) we get,

⇒ ㏒(b) = ㏒(a) + ㏒(c)/2 ⇒ (3).

⇒ ㏒(a)/x = ㏒(a) + ㏒(c)/2y = ㏒(c)/z.

⇒ ㏒(a)/x = ㏒(a) + ㏒(c)/2y and ㏒(a)/x = ㏒(c)/z.

Equate this equation, we get.

⇒ ㏒(a2y) = x㏒(a) + x㏒(c) and ㏒(a)/㏒(c) = x/z.

⇒ ㏒(a2y) - x㏒(a) = x㏒(c)  and ㏒(a)/(㏒(c) = x/z.

⇒ ㏒ a(2y - x) = x㏒(c)  and  ㏒(a)/㏒(c) = x/z.

⇒ ㏒(a)/㏒(c) = x/(2y - x)  and  ㏒(a)/㏒(c) = x/z.

⇒ x/(2y - x) = x/z.

⇒ 2y - x = z.

⇒ 2y = z + x.

This are the conditions of an Ap

Hence prove.

                                                                                           

MORE INFORMATION.

Some standard results,

(1) = Sum of first n natural number = ∑(r) = n(n + 1)/2.

(2) = Sum of first n odd natural number = ∑(2r - 1) = n².

(3) = Sum of first n even number = ∑(2r) = n(n + 1).

(4) = Sum of squares of first n natural numbers = ∑(r²) = n(n + 1)(2n + 1)/6.

(5) = Sum of cubes of first n natural numbers,

⇒ ∑(r³) = [n(n + 1)/2]² = [ ∑(r)]²

Answer 2

$\underline\blue{\bold{Given \: Question :- }}$

if a,b,c are in G.p and a^1/x= b^1/y= c^1/z then prove that x,y,z are in a.p.

____________________________________________

$\huge \orange{AηsωeR} ✍$

${ \boxed {\bf{Given}}}$

$\bf \:  ⟼ a, b, c \: are \: in \: GP$

$\bf \:  ⟼ {a}^{ \frac{1}{x} } = {b}^{ \frac{1}{y} } = {c}^{ \frac{1}{z} }$

${ \boxed {\bf{To \: Prove}}}$

$\bf \:  ⟼ x, y, z \: are \: in \: AP.$

${ \boxed {\bf{Solution}}}$

⟼ Since, a, b, c are in GP.

$\bf\implies \:\dfrac{b}{a} = \dfrac{c}{b}$

$\bf\implies \: {b}^{2} = ac\sf \: ⟼ \: (1)$

$\begin{gathered}\bf\red{Now,}\end{gathered}$

$\bf \:  ⟼ Consider \: {a}^{ \frac{1}{x} } = {b}^{ \frac{1}{y} } = {c}^{ \frac{1}{z} } = k$

$\bf\implies \: {a}^{ \frac{1}{x} } =k \: and \: {b}^{ \frac{1}{y} } = k \: and \: {c}^{ \frac{1}{z} } = k$

$\bf\implies \:a = {k}^{x} \sf \: ⟼ \: (2)$

$\bf\implies \:b = {k}^{y} \sf \: ⟼ \: (3)$

$\bf\implies \:c = {k}^{z} \sf \: ⟼ \: (4)$

⟼ On substituting equation (2), (3) and (4) in equation (1), we get

$\bf \:  ⟼ {k}^{2y} = {k}^{x} \times {k}^{z}$

$\bf \:  ⟼ {k}^{2y} = {k}^{x + z}$

$\bf\implies \:2y = x + z$

$\bf\implies \:y + y = x + z$

$\bf\implies \:y - x = z - y$

$\bf \:Hence, \: x, \: y, \: z \: are \: in \: AP$

___________________________________________

Geometric Progression (GP)

A sequence in which the ratio of two consecutive terms is constant is called geometric progression. The constant ratio is called common ratio(r).

$\bf \:  ⟼ r = \dfrac { { a }_{ n }+1 }{ { a }_{ n } }$