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

1 Answer
Alan P.
Dec 5, 2015

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)

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