Question

ABC are the AM between two numbers such that a + b + c = 15 and pqr be the HM between the same numbers such that 1/p + 1/q + 1/r = 5/3 then number are
(3,3)
(-1,-9)
(-3,-3)
(9,1) ​

Answer 1

Answer:

Let numbers be x,y

3(2x+y)=a+b+c=15

x+y=10

23(x1+y1)=(p1+q1+r1)=35

x1+y1=910

x=9,y=1

Hence, option 'D' is correct."(9, 1)" is correct option

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

Sum of the n terms of an arithmetic progression

$\displaystyle \: = \frac{n}{2} ( \: 1st \: \: term \: \: + \: last \: \: term \: \: )$

If a, b, c are the AM between two numbers x, y then

x, a, b, c, y are in AP

SO

$\displaystyle \: \frac{5}{2} (x + y) = x + a + b + c + y$

$\implies \: \displaystyle \: \frac{3}{2} (x + y) = a + b + c$

GIVEN

a , b, c are the AM between two numbers such that a + b + c = 15 and p , q , r be the HM between the same numbers such that

$\displaystyle \: \frac{1}{p} + \frac{1}{q} + \frac{1}{r} \: = \frac{5}{3}$

TO DETERMINE

The two numbers

EVALUATION

Let the two numbers are x, y

Since a, b, c are the AM between two numbers x, y

So

$\displaystyle \: \frac{3}{2} (x + y) = a + b + c \: \: \:$

$\implies \: \displaystyle \: \frac{3}{2} (x + y) = 15$

$x + y = 10 \: \: \: ....(1)$

Again

p , q , r be the HM between the numbers x, y

So

$\displaystyle \: \: \frac{1}{x} ,\frac{1}{p} ,\frac{1}{q} ,\frac{1}{r} ,\frac{1}{y} , \: \: are \: \: in \: AP$

So

$\displaystyle \: \frac{3}{2} ( \frac{1}{x} + \frac{1}{y} ) = \frac{1}{p} + \frac{1}{q} + \frac{1}{r} \: \: \: .......(1)$

$\implies \: \displaystyle \: \frac{3}{2} ( \frac{1}{x} + \frac{1}{y} ) = \frac{5}{3}$

$\implies \: \displaystyle \: ( \frac{1}{x} + \frac{1}{y} ) = \frac{10}{9}$

$\implies \: \displaystyle \: \frac{x + y}{xy} = \frac{10}{9}$

$\implies \: \displaystyle \: \frac{11}{xy} = \frac{10}{9}$

$\implies \: xy = 9 \: \: \: ....(2)$

From Equation (1) & (2) we get

$x(10 - x) = 9$

$\implies \: {x}^{2} - 10x + 9 = 0$

$\implies \: {x}^{2} -9x - x + 9 = 0$

$\implies \: x(x - 9) - 1(x - 9) = 0$

$\implies \: (x - 9)(x - 1) = 0$

Which gives x = 1 , 9

From Equation (1)

$x = 9 \: \: gives \: \: y = 1$

And

$x = 1 \: \: gives \: \: y = 9$

RESULT

So the required numbers are 9 , 1