Find k given that the following are consecutive terms of a geometric sequence: 7,k,28
if there are 3 terms in a geometric seqence, a,b,c, there is a common ratio,r, such that
ar=b
br=c
so
7r=k
rk=28
we can subsitute 7r for k
r(7r)=28
7r²=28
divide both sides by 7
r²=4
sqrt both sides
r=2 or -2 (pointed out by bcalle)
72=14
142=28
or
7-2=-14
-14-2=28
k is 14 or -14
If \(q\) is the ratio of the geometric progression:
\(q=\dfrac{7}{k}=\dfrac{k}{28}\iff \dfrac{7}{k}=\dfrac{k}{28}\iff k^2=196\iff\boxed{k=\pm14}\)