How do you find k such that k+1, 4k, 3k+5 is a geometric sequence?

1 Answer
Dec 5, 2015

Answer:

#k=1# or #k= -5/13#

Explanation:

If #k+1, 4k, 3k+5# is a geometric sequence
then the ratio between successive terms is equal.

#(k+1)/(4k) = (4k)/(3k+5)#

#rArr (k+1)(3k+5)=(4k)^2#

#rArr 3k^2+8k+5 = 16k^2#

#rArr 13k^2-8k-5=0#

We might be able to factor this directly or we could use the quadratic formula to determine the roots:
#color(white)("XXX")k= (8+-sqrt((-8)^2-4(13)(-5)))/(2(13)#

#color(white)("XXXX")= (8+-sqrt(324))/(2(13))#

#color(white)("XXXX")= (8+-sqrt(324))/(2(13)#

#color(white)("XXX")= (8+-18)/(2(13))#

#color(white)("XXXX")=(4+-9)/13#

#color(white)("XXXX")=13/13 = 1# or # = -5/13#

We could (and probably should) verify these results by checking that for each of these values of #k# the given sequence is geometric.

If #k=1#
then #k+1, 4k, 3k+5#
becomes #2, 4, 8# with an obvious common ratio of #2#

If #k=-5/13#
then #k+1, 4k, 3k+5#
becomes (with a little more effort) #8/13, -20/13, 50/13#
with a common ratio of #(-5/2)#