It is known that the first three numbers of four positive real numbers are equal difference sequence, the last three numbers are equal ratio sequence, the sum of the first and the third is 8, and the product of the second and the fourth is 36 (II) if the first three numbers are the first three terms of the arithmetic sequence an and the last three numbers are the first three terms of the proportional sequence BN, let CN= an.bn To find the first n terms and TN of the sequence CN

It is known that the first three numbers of four positive real numbers are equal difference sequence, the last three numbers are equal ratio sequence, the sum of the first and the third is 8, and the product of the second and the fourth is 36 (II) if the first three numbers are the first three terms of the arithmetic sequence an and the last three numbers are the first three terms of the proportional sequence BN, let CN= an.bn To find the first n terms and TN of the sequence CN

1）a2=(a1+a3)/2=8/2=4 a4=36／a2=9
a3=√﹙a2a4﹚=√36=6
a1=8-a3=2
2)an=2n
bn=4×﹙3/2﹚^(n-1)
cn=8n×﹙3/2﹚^(n-1)
Tn/8=1＋2×3/2＋3×9/4＋…… ＋n×﹙3/2﹚^(n-1)
3/16Tn=3/2＋2×9/4＋…… ＋(n-1)×﹙3/2﹚^(n-1)＋n×﹙3/2﹚^n
-1/16Tn=1＋3/2＋9/4＋…… ＋﹙3/2﹚^(n-1)-n×﹙3/2﹚^n
=[1-﹙3/2﹚^n]/(1-3/2)－n×(3/2)^n
=(2-n)(3/2)^n－2
Tn=(n-2)3^n/2^(n-4)＋32