Find k

Find k so that the terms k-3, k+1, and 4k-2 form a geometric sequence. Show your solution.

Correct answer:

k1 =  5
k2 =  1/3

Step-by-step explanation:

g1=k3 g2=k+1 g3 = 4k2  q = q2/q1= q3/q2  q22 = q1 q3  (k+1)2=(k3) (4 k2)  (k+1)2=(k3) (4 k2) 3k2+16k5=0 3k216k+5=0  a=3;b=16;c=5 D=b24ac=162435=196 D>0  k1,2=2ab±D=616±196 k1,2=616±14 k1,2=2.666667±2.333333 k1=5 k2=0.333333333   Verifying Solution:  g1=k13=53=2 g2=k1+1=5+1=6 g3=4 k12=4 52=18 r1=g2/g1=6/2=3 r2=g3/g2=18/6=3

Our quadratic equation calculator calculates it.

G1=k23=0.33333=38=2322.6667 G2=k2+1=0.3333+1=34=1311.3333 G3=4 k22=4 0.33332=320.6667 r2=G2/G1=1.3333/(2.6667)=21=0.5 R2=G3/G2=(0.6667)/1.3333=21=0.5  k2=0.3333=31

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