It is known that the equal ratio sequence an is an increasing sequence and the square of A5 = A102 (an + an + 2) = 5An + 1 to find the general term formula A5 ^ 2 = a10. Get (A1 * q ^ 4) ^ 2 = A1 * q ^ 9 get A1 = Q 2 {an + a (n + 2)} = 5An + 1 leads to 2 (1 + Q ^ 2) = 5q leads to q = 1 / 2 or 2 So A1 = q = 2 an=2^n 2 {an + a (n + 2)} = 5An + 1 how to get 2 (1 + Q ^ 2) = 5q

It is known that the equal ratio sequence an is an increasing sequence and the square of A5 = A102 (an + an + 2) = 5An + 1 to find the general term formula A5 ^ 2 = a10. Get (A1 * q ^ 4) ^ 2 = A1 * q ^ 9 get A1 = Q 2 {an + a (n + 2)} = 5An + 1 leads to 2 (1 + Q ^ 2) = 5q leads to q = 1 / 2 or 2 So A1 = q = 2 an=2^n 2 {an + a (n + 2)} = 5An + 1 how to get 2 (1 + Q ^ 2) = 5q

A 5 ^ 2 = a 10. Get (a 1 * q ^ 4) ^ 2 = a 1 * q ^ 9, get a 1 = QaN as increasing sequence, explain Q > 1 2 [an + a (n + 2)] = 5A (n + 1) a (n + 2) = an · Q ^ 2; a (n + 1) = an · Q ^ 2. Substitute the above formula to get: 2An (1 + Q ^ 2) = 5An · Q, cancel an about, then get: 2 (1 + Q ^ 2) = 5q (2q-1) (Q-2) = 0q = 0.5 (rounding) or q = 2

In the equal ratio sequence an, a1 + A2 + a3 = 6, A4 + A5 + A6 = 48 are known, and the general term formula of the sequence is obtained

a4/a1=a5/a2=a6/a3=q³
So Q & # 179; = (A4 + A5 + A6) / (a1 + A2 + a3) = 8
q=2
a1+a2+a3
=a1+a1q+a1q²
=a1(1+q+q²)=6
So A1 = 6 / 7
So an = 6 / 7 * 2 ^ (n-1)

Kneel down and beg Let the sum of the first n terms of {an} be the general term formula of {an} if Sn = 242

Because an is an equal ratio sequence, A5 = A2 * q ^ 3 162 = 6q ^ 3
q=3
a1=a2/q=6/3=2
an=a1q^(n-1)=6*3^(n-1)=2*3^n
sn=a1(1-q^n)/(1-q)
242=2*(1-3^n)/(1-3)
n=5

In the known equal ratio sequence {an}, A2 = 6, A5 = 162, find the general term formula of the sequence {an}? If the first n terms and Sn of the sequence {an} are more than 50, find the minimum value of N?

162 / 6 = 27, so the common ratio is 3. Because 3 ^ 3 = 27, the first term is 6 / 3 = 2, so if the general term formula is 2 * 3 ^ (n-1) Sn > 50, then 2 (3 ^ n-1) / (3-1) > 50 can calculate n > 3. Therefore, the minimum n is 4