As (k-8), (k+4) and (3k+2) are in geometric series
(k+4)^2=(k-8)(3k+2)
or k^2+8k+16=3k^2-22k-16
or 2k^2-30k-32=0
or k^2-15k-16=0
or (k-16)(k+1)=0
i.e. k=16 or k=-1
Hence, theuir could be two geometric series
(1) If k=16, first three terms are 8,20 and 50 i.e. first term is 8 and common ratio is 20/8=5/2 and sixth term is 8*(5/2)^5=3125/4 and sum of first 10 terms is 8((5/2)^10-1)/(5/2-1)=8(9765625/1024-1)xx2/3=8xx9764601/1024xx2/3=3254867/64=50857 19/64
(2) If k=-1, first three terms are -9,3 and -1 i.e. first term is -9 and common ratio is 3/(-9)=-1/3 and sixth term is (-9)*(-1/3)^5=1/27 and sum of first 10 terms is (-9)((-1/3)^10-1)/(-1/3-1)=9(1/59049-1)xx3/4=-27/4xx59048/59049=-14762/2187=-6 1640/2187
