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#