Question

if a^2 , b^2, c^2 is in Ap , prove that a/ b+c ,
b/ c+a, c/ a+b .
plz solve this question and explain by pic also nofreepointshere ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\: {a}^{2}, \: {b}^{2}, \: {c}^{2} \: are \: in \: AP$

$\bf\implies \:\boxed{ \tt{ \: {b}^{2} - {a}^{2} = {c}^{2} - {b}^{2}}} - - - (1)$

Now,

$\rm :\longmapsto\:\dfrac{a}{b + c}, \: \dfrac{b}{c + a}, \: \dfrac{c}{a + b} \: are \: in \: AP$

Adding 1 in each term,

$\rm :\longmapsto\:if \: \dfrac{a}{b + c} + 1, \: \dfrac{b}{c + a} + 1, \: \dfrac{c}{a + b} + 1 \: are \: in \: AP$

$\rm :\longmapsto\:if \: \dfrac{a + b + c}{b + c}, \: \dfrac{b + c + a}{c + a}, \: \dfrac{c + a + b}{a + b}\: are \: in \: AP$

On divide by a + b + c, we get

$\rm :\longmapsto\:if \: \dfrac{1}{b + c}, \: \dfrac{1}{c + a}, \: \dfrac{1}{a + b} \: are \: in \: AP$

$\rm :\longmapsto\:if \: \dfrac{1}{c + a} - \dfrac{1}{b + c} = \dfrac{1}{a + b} - \dfrac{1}{c + a}$

$\rm :\longmapsto\:if \: \dfrac{b + c - c - a}{(c + a)(b + c)} = \dfrac{c + a - a - b}{(a + b)(c + a)}$

$\rm :\longmapsto\:if \: \dfrac{b - a}{b + c} = \dfrac{c - b}{a + b}$

$\rm :\longmapsto\:if \: (b + c)(c - b) = (b - a)(a + b)$

$\rm :\longmapsto\:if \: {c}^{2} - {b}^{2} = {b}^{2} - {a}^{2}$

$\rm :\longmapsto\: {a}^{2}, \: {b}^{2}, \: {c}^{2} \: are \: in \: AP$

Hence, Proved

More to know :-

1. If each term of an AP series is increased by a constant quantity, the resulting new series is also an AP series.

2. If each term of an AP series is decreased by a constant quantity, the resulting new series is also an AP series.

3. If each term of an AP series is multiplied by a constant quantity, the resulting new series is also an AP series.

4. If each term of an AP series is divided by a non zero constant quantity, the resulting new series is also an AP series.

Answer 2

Answer:

Given that,

\rm :\longmapsto\: {a}^{2}, \: {b}^{2}, \: {c}^{2} \: are \: in \: AP:⟼a

2

,b

2

,c

2

areinAP

\bf\implies \:\boxed{ \tt{ \: {b}^{2} - {a}^{2} = {c}^{2} - {b}^{2}}} - - - (1)⟹

b

2

−a

2

=c

2

−b

2

−−−(1)

Now,

\rm :\longmapsto\:\dfrac{a}{b + c}, \: \dfrac{b}{c + a}, \: \dfrac{c}{a + b} \: are \: in \: AP:⟼

b+c

a

,

c+a

b

,

a+b

c

areinAP

Adding 1 in each term,

\rm :\longmapsto\:if \: \dfrac{a}{b + c} + 1, \: \dfrac{b}{c + a} + 1, \: \dfrac{c}{a + b} + 1 \: are \: in \: AP:⟼if

b+c

a

+1,

c+a

b

+1,

a+b

c

+1areinAP

\rm :\longmapsto\:if \: \dfrac{a + b + c}{b + c}, \: \dfrac{b + c + a}{c + a}, \: \dfrac{c + a + b}{a + b}\: are \: in \: AP:⟼if

b+c

a+b+c

,

c+a

b+c+a

,

a+b

c+a+b

areinAP

On divide by a + b + c, we get

\rm :\longmapsto\:if \: \dfrac{1}{b + c}, \: \dfrac{1}{c + a}, \: \dfrac{1}{a + b} \: are \: in \: AP:⟼if

b+c

1

,

c+a

1

,

a+b

1

areinAP

\rm :\longmapsto\:if \: \dfrac{1}{c + a} - \dfrac{1}{b + c} = \dfrac{1}{a + b} - \dfrac{1}{c + a}:⟼if

c+a

1

−

b+c

1

=

a+b

1

−

c+a

1

\rm :\longmapsto\:if \: \dfrac{b + c - c - a}{(c + a)(b + c)} = \dfrac{c + a - a - b}{(a + b)(c + a)}:⟼if

(c+a)(b+c)

b+c−c−a

=

(a+b)(c+a)

c+a−a−b

\rm :\longmapsto\:if \: \dfrac{b - a}{b + c} = \dfrac{c - b}{a + b}:⟼if b+cb−a = a+bc−b\rm :\longmapsto\:if \: (b + c)(c - b) = (b - a)(a + b):⟼if(b+c)(c−b)=(b−a)(a+b)\rm :\longmapsto\:if \: {c}^{2} - {b}^{2} = {b}^{2} - {a^{2}:⟼ifc 2 −b 2 =b2 −a 2\rm :\longmapsto\: {a}^{2}, \: {b}^{2}, \: {c}^{2} \: are \: in \: AP:⟼a 2 ,b 2c2 areinAP

Hence, Proved

More to know :-

1. If each term of an AP series is increased by a constant quantity, the resulting new series is also an AP series.

2. If each term of an AP series is decreased by a constant quantity, the resulting new series is also an AP series.

3. If each term of an AP series is multiplied by a constant quantity, the resulting new series is also an AP series.

4. If each term of an AP series is divided by a non zero constant quantity, the resulting new series is also an AP series.