Question

if the product of three no in gp be 216and their sum is 19 then the no are​

Answer 1

Given:

Product of three numbers in G.P. = 216

Sum of three numbers in G.P. = 19

Answer:

$a=6$

$r= 1.5 \: \: \: or \: \: \: r = 0.6667$

Explanation:

$Let \: \: the \: \: three \: \: numbers \: \: be \: \: a \:, \: \frac{a}{r} \: , \: ar$

$a \times \frac{a}{r} \times ar = 216.............(i)$

$a + \frac{a}{r} + ar = 19.............(ii)$

ATQ,

$a \times \frac{a}{r} \times ar = 216$

Cancelling r and multiplying, we get:

${a}^{3} = 216$

Cube rooting both sides , we get:

$a = 6$

Substituting the value of a in equation (ii) , we get:

$6 + \frac{6}{r} + 6r = 19$

Taking LCM of LHS, we get:

$\frac{6r + 6 + 6 {r}^{2} }{r} = 19$

Cross multiplying, we get:

$6r + 6 + 6 {r}^{2} = 19r$

Sending 19r to RHS, we get:

$6 {r}^{2} - 13r + 6 = 0$

Using Sridharacharya Method or Quadratic Formula,

Sridharacharya Method or Quadratic Formula

$x = \frac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac } }{2a}$

where,

x= roots of the quadratic equation

a= constant in ax²+bx+c

b= constant in ax²+bx+c

c= constant in ax²+bx+c

Here,

x= we have to calculate

a=6

b=-13

c=6

Substituting the value of a, b and c in the above formula, we get:

$r = \frac{ - ( - 13) \:\pm \: \sqrt{ {( - 13)}^{2} - 4 \times 6 \times 6 } }{2 \times 6}$

$r = \frac{13 \: \pm \: \sqrt{169 - 144} }{12}$

$r = \frac{13 \: \pm \: \sqrt{25} }{12}$

We know that √25=5

$r = \frac{ 13 \: \pm \: 5}{12}$

$r = \frac{13 + 5}{12} \: \: or \: \: r = \frac{13 - 5}{12}$

$r = \frac{18}{12} \: \: or \: \: r = \frac{8}{12}$

Simplifying the Fraction, we get:

$r = \frac{3}{2} \: \: or \: \: r = \frac{2}{3}$

$r = 1.5 \: \: or \: \: r = 0.66667$

Additional Information:

GP is Geometric Progression.

In G.P. ,the terms are a,ar, ar²,and so on.

The difference between two terms is $\dfrac{a}{r}$.