It is known that a3 3 2, S3 9 2 find a1 and q

q = -1/2 a1 = a3/q^2 =6

The process of solving the problem is as follows:

When q=1, it is clear that a1 = a2 =a3 = 3 2, s3=9 2, satisfies the requirements;

When q is not equal to 1, a3 = a1*q 2 = 3 2 , s3 = a1*(1-q 3) (1-q) = 9 2; Divide by two formulas to eliminate a1 and get.

q 2(1-q) (1-q 3)=q 2(1-q) (1-q)(1+q+q 2)=q 2 (1+q+q 2)1 3, then it is simplified as.

2q 2-q-1=0, factorization.

(q-1)(2q+1)=0, excluding q=1We get q = -1 2, at which point, a1 = a3 q 2 =6

q = 1/2 a1 = a3/q^2 =6

The process of solving the problem is as follows:

When q=1, it is clear that a1 = a2 =a3 = 3 2, s3=9 2, satisfies the requirements;

When q is not equal to 1, a3 = a1*q 2 = 3 2 , s3 = a1*(1-q 3) (1-q) = 9 2; The two-formula phase code is removed to eliminate a1 and obtain.

q 2(1-q) (1-q 3)=q 2(1-q) (1-q)(1+q+q 2)=q 2 (1+q+q 2)1 3, then it is simplified as.

2q 2-q-1=0, factorized to (q-1)(2q+1)=0, excluding q=1We get q = -1 2, at which point, a1 = a3 q 2 =6

This is to find the first term and the common ratio of the common ratio series.

Let the first term of the series be a1 and the common ratio be q, then.

a3=aq^2=3/2

s3=a1+a2+a3=a1+a1q+3 2=9 2 Solve the above equations to get a1=3 2,q=1 or a1=6,q=-1 2Please pay attention to [my rate] first

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a3=a1q^2=3/2

s3=a1+a1q+a1q 2=9 2q 2 (1+q+q 2)=1 Bi Cover Hand 3

2q^2-q-1=0

2q+1)(q-1)=0

q=-1 remorse2, a1=(3 2) q 2=6 or: q=1, a1=3 2

in proportional series.

, (1) a1=2, s3=26, find q and a3 formula: s3=a1+a2+a3 a2=a1*q a3=a1*q 2 then s3=a1+a1*q+a1*q 2

Bringing a1=2 and s3=26 into the Pipe upper volume grip group, we can get: 26=2+2q+2q*q

So: q=-4 or state orange q=3, then a3=32 or 18

a9=a1*q^8

So q 8 = (1 243) 27 = 1 3 8q<0

So q=-1 3

s5=a1*(1-q^5)/(1-q)=27*(1-1/3^5)/(1+1/3)=121/6

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Summary. You can shoot the question.

Known =q=3 to find a

You can shoot the question.

Hold on. Because a5 a1q is 81 to the 4th power, and because q 3, then there is a1x3 to the fourth power 81, and since a1 81 81, 3 is 81 to the fourth power

a3=a1q^2=3/2

s3=a1+a1q+a1q^2=9/2

q 2 (1+q+q 2)=1 32q 2-q-1=0

2q+1)(q-1)=0

q=-1 2, a1=(3 slag 2) q 2=6 or: q=1, a1=3 2

q = -1/2 a1 = a3/q^2 =6

The process of solving the problem is as follows:

When q=1, it is clear that a1 = a2 =a3 = 3 2, s3=9 2, satisfies the requirements;

When q is not equal to 1, a3 = a1*q 2 = 3 2 , s3 = a1*(1-q 3) (1-q) = 9 2; Divide by two formulas to eliminate a1 and get.

q 2(1-q) (1-q 3)=q 2(1-q) (1-q)(1+q+q 2)=q 2 (1+q+q 2)1 3, then it is simplified as.

2q 2-q-1=0, factorized to (q-1)(2q+1)=0, excluding q=1We get q = -1 2, at which point, a1 = a3 q 2 =6

Proportional series properties.

1) If m, n, p, q n*, and m+n=p+q, then am*an=ap*aq.

2) In the proportional series, the sum of each k terms in turn is still a proportional series.

3) If "g is the proportional middle term of a and b", then "g 2=ab(g≠0)".

4) If it is a proportional series, the common ratio is q1, which is also a proportional series, and the common ratio is q2, then it is ,...is a proportional series, the common ratio is q1 2, q1 3...., c is a constant, , is a series of proportional numbers, and the common ratio is q1, q1q2, q1 q2.

5) If (an) is a proportional sequence and each item is positive, and the common ratio is q, then (log, with a as the base an, and the logarithm of an) as the difference, and the tolerance is the logarithm of log with a as the base q.

According to the title, when Q=1, it is obvious that A1 = A2 =A3 = 3 2, S3=9 2, satisfies the requirements;

When q is not equal to 1, a3 = a1*q 2 = 3 2 , s3 = a1*(1-q 3) (1-q) = 9 2; Divide by two formulas to eliminate a1 and get.

q 2(1-q) (1-q 3)=q 2(1-q) (1-q)(1+q+q 2)=q 2 (1+q+q 2)1 3, then it is simplified as.

2q 2-q-1=0, factorized to (q-1)(2q+1)=0, excluding q=1We get q = -1 2, at which point, a1 = a3 q 2 =6

s3-a3=a1+a2=a1+a1*q=2

a3=a1*q*q=3/2

Solving the system of equations will do it.