It is known that the tolerance of the arithmetic sequence {an} is greater than 0, and A3 and A5 are two of the equations X & sup2; - 14x + 45 = 0. The sum of the first n terms of the sequence {BN} is Sn = 1-1 / 2B (1) Finding the general term formula of sequence {an}, {BN}; (2) Let CN = anbn prove CN + 1 ≤ cn

It is known that the tolerance of the arithmetic sequence {an} is greater than 0, and A3 and A5 are two of the equations X & sup2; - 14x + 45 = 0. The sum of the first n terms of the sequence {BN} is Sn = 1-1 / 2B (1) Finding the general term formula of sequence {an}, {BN}; (2) Let CN = anbn prove CN + 1 ≤ cn

x²-14x+45=0
(x-5)(x-9)=0
X1=5 x2=9
Because the tolerance is greater than 0
So A3 = 5, A5 = 9
a3+2d=a5
5+2d=9
d=2
a1=a3-2d=5-4=1
an=a1+(n-1)d=1+2(n-1)=2n-1

Let {an} be an arithmetic sequence with a tolerance of - 2 if a1 + A4 + A7 + +A97 = 50, then A3 + A6 + A9 + +A99 equals ()
A. 82B. -82C. 132D. -132

Because {an} is an arithmetic sequence with tolerance of - 2, A3 + A6 + A9 + + A99 = (a1 + 2D) + (A4 + 2D) + (A7 + 2D) + +(A97 + 2D) = a1 + A4 + A7 + + A97 + 33 × 2D = 50-132 = - 82

Let {an} be an arithmetic sequence with a tolerance of - 2 if a1 + A4 + A7 + +A97 = 50, then A3 + A6 + A9 + +A99 equals ()
A. 82B. -82C. 132D. -132

Because {an} is an arithmetic sequence with tolerance of - 2, A3 + A6 + A9 + + A99 = (a1 + 2D) + (A4 + 2D) + (A7 + 2D) + +(A97 + 2D) = a1 + A4 + A7 + + A97 + 33 × 2D = 50-132 = - 82