Question

product of 3 numbers in G.P = 216 and the sum of their squares = 189 Then one of the numbers of this set can be?

Answer 1

Let the three numbers be a/r , a , ar

Answer 2

a/r * a * ar = a^3 = 216

a = 6

Answer 3

thats alright,,what eqn you made next?

Answer 4

a^2/r^2 + a^2 + a^2 r^2 = 189

Answer 5

so?? what next/?

Answer 6

put the value of a ... the equation is Quadratic in r^2

Answer 7

36 r^4 + 36 = 153r^2

Answer 8

solve for r^2

Answer 9

36(r^4 + 1) = 153 r^2

r^4 + 1 = 4.25 r^2

Answer 10

let r^2 = y

y^2 + 1 = 4.25 y

Answer 11

y^2 - 4.25y + 1 =0

y = 4 y = 0.5

Answer 12

@experimentX any other easy way to solve this

Answer 13

this is the easiest way!!

Answer 14

ok

Answer 15

r = 2

Answer 16

6 * 2 = 12

Answer 17

very smart, easier i think than the idea of starting with \(a, ar, ar^2\) and then \(a^3r^2=216\)

Answer 18

well if you let 3 nos. to be a,b,c, we have
abc = 216
ac=b^2
a^2 + b^2 +c^2 =189
from 1st and 2nd, b=6
and from 3rd,
a^2 + c^2 + ac = 189
we know ac=36 from 1st eqn since b=6
(a+c)^2 - ac = 189
and (a-c)^2 + 3ac = 189
from these 2 eqns,
a+c = 15
and
a-c = 9

a= 12 and c=3
so nos are 12,6 and 3

Answer 19

this is lot more simple thxxx