Question

. If (b+c), (c+a), (a+b) are in HP, then a^2, b^2, c^2 are in??
a) AP
b) GP
c) HP
d) None​

Answer 1

Answer:

a) A.P

Step-by-step explanation:

Given,

(b + c) , (c + a) , (a + b) are in H.P

To Find :-

a² , b² , c² are in

How To Find :-

By using given condition we need to write them as reciprocals of elements of Arithmetic Progression. After doing that and simplifying that we will get an equation on a² , b² , c² .  We need to check that whether they are in A.P or G.P or H.P.

Formula Required :-

If x , y , z are in H.P then :-

$\dfrac{1}{y}-\dfrac{1}{x}=\dfrac{1}{z}-\dfrac{1}{y}$

$\dfrac{1}{y}+\dfrac{1}{y}=\dfrac{1}{z}+\dfrac{1}{x}$

$\dfrac{2}{y}=\dfrac{x+z}{xz}$

$\implies y=\dfrac{2xz}{x+z}$

∴ If x , y , z are in H.P then 'y' = 2xz/x + z.

If s , t , u are in A.P then :-

t - s = u - t

t + t = u + s

2t = u + s

∴ If s , t , u are in A.P then '2t = u + s'

Solution :-

(b + c) , (c + a) , (a + b) are in H.P :-

$c+a=\dfrac{2\times (b+c)\times (a+b)}{b+c+a+b}$

$c+a=\dfrac{(2b + 2c)\times (a + b)}{a+2b+c}$

$(c+a)\times (a+2b+c)=a(2b+2c)+b(2b+2c)$

$c(a+2b+c)+a(a+2b+c)=2ab+2ac+2b^2+2bc$

ac + 2bc + c² + a² + 2ab + ac = 2ab + 2ac + 2b² + 2bc

2ac + 2bc + a² + c² + 2ab = 2ab + 2ac + 2b² + 2bc

2ac + 3bc + 2ab - 2ab - 2ac - 2bc + a² + c² = 2b²

a² + c² = 2b²

a² - b² = b² - c²

∴ a² , b² , c² are in A.P