In {an}, an + 1 > an, a1 + A4 = 9, A2 * A3 = 8, S10 =?

In {an}, an + 1 > an, a1 + A4 = 9, A2 * A3 = 8, S10 =?

a4=a1*q^3 a1+a1*q^3=9 (1)
A2 = A1 * q A3 = A1 * q ^ 2 A1 ^ 2 * q ^ 3 = 8 (2) is solved by substituting equation (1) (2)
A1 = 8 or A1 = 1
So we can get Q1 = 1 / 2 or Q1 = 2, because an + 1 > an, so we take A1 = 1, Q1 = 2
s10=2^10-1

If A1 = 1 / 2, A4 = 4, then q =? A1 + A2 + +an=?

In an, A1 = 1 / 2, A4 = 4
Then the common ratio q = (A4 / A1) to the third power = 8 to the third power = 2
a1+a2+… +an=Sn=a1(1-q^n)/(1-q)=1/2（1-2^n)/(1-2)=2^(n-1) -1/2

If A1 = 8, A4 = 64, then the common ratio q is ()
A. 2B. 3C. 4D. 8

From a4a1 = Q3 = 648 = 8, q = 2

In the equal ratio sequence an, the common ratio q is greater than 1, and a1 + A4 = 9a2xa3 = 8

a1+a4=9
a2xa3=8
a2*a3=a1*a4
therefore
a1+a4=9
a1*a4=8
The solution is: A1 = 1, A4 = 8, q = 2
Or A1 = 8 A4 = 1 Q = 1 / 2 (rounding)
So the first term of the sequence A1 = 1, the common ratio q = 2