Question

find the value of k so that 2k+2, 5k-11, and 7k-13 is a geometric sequence.

Answer 1

Can you find common ratio here?

Answer 2

i think k is the common ratio

Answer 3

common ratio = 5k-11 / 2k+ 2 and this equals 7k-13 / 5k-11

solve this for k

Answer 4

it didn't say those were consecutive terms though

Answer 5

......common ratio in = (b)/ (a) where b and a are in geometric sequence

Answer 6

@lgbasallote ..no

Answer 7

@mathslover that only applies when b and a are consecutive

Answer 8

@babybaby what do you mean no?

Answer 9

it was not mentioned in my book that it is a geometric sequence. sorry

Answer 10

yes thats true of course - i assumed they were
how else could we solve this

Answer 11

...so it's not geometric sequence?

Answer 12

lol

Answer 13

im sorry... what i mean is that it is not mentioned as consecutive

Answer 14

this just got interesting

Answer 15

Can we apply the sum of geometric sequence formula and then equating?

Answer 16

if they are consecutive
equation reduces to 11k^2 - 98k + 147 = 0

Answer 17

i can't think of a way to do this unless they are consecutive terms

Answer 18

........

Answer 19

@ mathslover .
yes

Answer 20

lets try if it is a consecutive terms

Answer 21

k=7

Answer 22

@Algebraic! how did you answer it?

Answer 23

same way as @cwrw238 evidently :)

Answer 24

solving the equation in last post gives k = 7

Answer 25

set up 3 eqn.:

a(r)^(n-1) = 2k+2
a(r)^n =5k -11
a*r^(n+1) = 7k-13

Answer 26

http://www.wolframalpha.com/input/?i=11k%5E2+-+98k+%2B+147+%3D+0

Answer 27

how it was reduced?

Answer 28

also another value is 21/11

Answer 29

start solving stuff in terms of k... eg. r= (5k-11)/(2x+2)

Answer 30

@cwrw238 I saw that but I didn't check to make sure it worked... it does I guess..

Answer 31

i knew it.

Answer 32

thanks you guys. . .

Answer 33

7 gives a gp 16 , 24 , 36

Answer 34

@cwrw238 aye

Answer 35

yw