Question

given that x1, x2, x3,......x16 are positive real numbers such that
\[\frac{ x1 }{ x2 }=\frac{ x2 }{ x3 }=\frac{ x3 }{ x4 }=....\frac{ x15 }{ x16}\]
if x1+x2+x3+x4=20
x5+x6+x7+x8=320,
then what is x13+x14+x15+x16 ?

Answer 1

@hartnn

Answer 2

this is not so difficult

let x1/x2 = ..... = r

Answer 3

ohk ..,,den .?

Answer 4

so, x1 = r x2 = r^2 x3 = r^3 x4 = ....and so on

so from
x1+x2+x3+x4=20

and
x5+x6+x7+x8=320

you should be able to find 'r'
try it.

Answer 5

hm but how r^2 and r^3 came ..?

Answer 6

x2/x3 = r

x2 = r x3

x1 = r x2 = r (r x3) = r^2 x3

Answer 7

got got it ! r yes

Answer 8

x1+x2+x3+x4=20

r^4 x5 + r^3 x5 + r^2 x5 + r x5 = 20

x5 (r^4+r^3+r^2+r) = 20

Answer 9

umm ..i did this like
\[x1(1+\frac{ 1 }{ r }+\frac{ 1 }{ r^2 }+\frac{ 1 }{ r^3 } ) = 20\]

Answer 10

something like G.P

Answer 11

no prob :)
see whether you get r as 1/2 or not.

Answer 12

they are surely in GP.

Answer 13

yes ./.. wow ..r=1/2 ..got that

Answer 14

now find x1 from your last equation.

Answer 15

then you can find any term,
x1 = r x2 = r^2 x3 = r^3 x4 = ....and so on

x13 = r^12 x1
x14 = r^13 x1
x15 = r^14 x1
x16 = r^15 x1

Answer 16

and then again using ..that by converting and putting x1 and r ..in what we need to find .!

Answer 17

x1 comes out to be 5/3 .??

Answer 18

4/3

Answer 19

x1 (1+2+4+8) =20

x1 *15 = 20

Answer 20

hahahah ..lol ...20/15 i wrote ..5/3 ..mad i am

Answer 21

answer is absolutely ..correct ..thank u

Answer 22

welcome ^_^