In the first year of high school, in the proportional number series an, a1 a5 15, a4 a2 6, find a3.

a1-(a1+4d)=15

a1-(a1+d)=6

Calculate a1 and d, then a3=a1+2d

Just bring in the data, it should be easy to figure it out, and I still have to do my homework, so you can do the math yourself.

Solution: It can be obtained from the meaning of the question:

a1-a5=a1-a1q^4=15

a4-a2=a1q^3-a1q=6

Comparison: 1-q 2) q = -5 2

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

So q=-1 2 or q=-2

So a1 = 16 or a1 = -1

So a3 = a1q 2 = 4 or -4

Solution: Let the ratio be m, then.

a1=m a5,a2=m a5,a3=m a5,a4=ma5a1-a5=(m -1)a5=15, if a5=1, m -1=15, get m = 2, if a, m has no solution.

again a4-a2=m-m =6, and the equation holds only when m=-2.

a3=m²a5=（-2）²×1=4

a1=a3/q^2

a5=a3*q^2

a4=a3q

a2=a3/q

The answer can be found in the substitution.

Get q = -1 2

a3=4

a1-a1q^4=15 (1)

a1q^3-a1q=6 (2)

1) (2) There are only two unknowns of A1 and Q in the two formulas, which can be solved.

After the solution, there is a general term formula, and A3 can naturally be found!

Because of the proportional series, a1, a2, a4, and a5 are all represented by a3 and q, that is, a1=a3 q, a2=a3 q, a4=a3q, a5=a3q

According to the known a1 a5 = 15 i.e. a3 q a3q = 15, a4 a2 = 6 i.e. a3q a3 q = 6, the solution is a3 = 4 or -4, q = 1 2 or -2

Hope you're satisfied!

The proportional number of Dou Xinhong column blank book has a5 + a7 = (a1 + a3) * q 4

q 4=4 Tan socks 8=1 2

a9+a11+a13+a15=(a1+a3+a5+a7)*q^8=(8+4)*1/4=3

Proportional sequence of the general term formula an=a1 q (n-1)a5-a1=a1 q 4-a1=15a4-a2=a1 q 3-a1 q 2=6 solution equation q=2 or q=1 2q=2,a1(q 4-1)=15,a1=1,a3=a1q 2=4q=1 2,a1(q 4-1)=15,a1=-16,a3=a1q 2=-4, so a3=4 or -4

A3 is equal to the square of A1 Q, and A4 is equal to the square of A2 Q, so A3+A4 divided by A1+A2 is also equal to the square of Q, so 40 divided by 10 is inherited from 4, and 4 is the square of Q. In the same way, a5 + a6 divided by a1 + a2 is equal to q to the 4th power, q is squared to the 4th, q is 4, q is 16 times to grasp the Zen Bifang, so a5 + a6 is equal to 10 16 = 160

According to the nature of the proportional series, it is accompanied by the year.

A1 + A2, A3 + A4, A5 + A6 are proportional series, and the number of stools is messy.

a3 + a4) = a1 + a2) (a5 + a6), i.e. 40 potato brigade = 10 (a5 + a6).

The solution yields a5+a6=160

The answer is three-quarters, because (a2+a3+a4) (a1+a2+a3)=q(-1 2), so a3+.a8 = q square x (a1 + a2 + a3 + a2 + a3 + a4) = 3 4

a2=a1*q, a5=a1*q 4, q=1 2, a1=4, so an=2 (3-n).

So anan+1=2 (3-n)*2 (2-n)=2*4 (2-n) is a proportional series with the first term 8 and the common ratio of 1 4.

Therefore, the original formula = 8[1-(1 4) n] [1-(1 4)], when n tends to infinity, (1 4) n tends to 0, so the value range of the original formula is [8,32 3).

into GP, so Q 3=A5 A2=1 8 so Q=1 2An=4 1 2 (N-1)=2 (3-N) Let An*A(N+1)=bn=2 (3-N)*2 (2-N)=2 (5-2N) The sum of the first n terms is set to TN

tn=2^3+2^1+2^(-1)+2^(-3)+…2^(5-2n) ①

tn/4=2^1+2^(-1)+2^(-3)+…2^(5-2n)+2^(3-2n) ②

Yes: 3tn 4=8-2 (3-2n)=8-(8 4 n) so tn=32 3(1-1 4 n) choose c

q^3=a5/a2=1/8

q=1/2a1=4

Let bn=ana(n+1).

Then bn is also proportional.

b1=8q'=(a2a3) (a1a2)=a3 a1=q 2=1 4, so =8*(1-1 4 n) (1-1 4) choose c

Answer: c. Because a1a2, a2a3, a3a4....It also forms an equal ratio series, and it is good to use the proportional summation formula.

c let a0=8, let the original formula = t

Then 4t=a0a1+a1a2+.an-1an

Subtract the two formulas to get 3t, and then divide by 3.

Let the common ratio be q, and the original equation will be.

a1(q^4-1)=15,①

a1(q^3-q)=6,②

, (q 2 + 1) q = 5 2, the solution is q = 2 or 1 2, which is substituted into , a1 = 1, or -16, a3 = a1q 2 = 4 or -4

Solution: is a proportional series.

Let the first term be A1 and the common ratio is Q

a5=a1×q×q×q×q，a4=a1q×q×q，a2=a1×q，a1=a1

It is known that a5-a1=15, a4-a2=6 (a5-a1) (a4-a2)=5 2

q+1=5 2, q=2 or 1 2, a1=1 or -16 a3=4 or -4

a5=a1q^4 ， a4=a1q^3 ， a2=a1qa1q^4-a1=15 ；a1q 3-a1q=6a1(q 4-1)=15 (a) ; a1(q 3-q) = 6 (b).

a1(q^4-1)/a1(q^3-q)=5/2q^4-1/q^3-q=5/2 ( q^2-1)(q^2+1)/q(q^2-1)=5/2 (q^2+1)/q=5/2

2q^2+2=5q

q=1 2 or 2

Bring in a quest for a3 respectively.

Ask for the rest.

A3q 2-a3 (q 2)=15 from a5-a1=15 and a3q-a3 q=6 from a4-a2=6

From:

q^2-1/(q^2)]/(q-1/q)=5/2q+1/q=5/2

2q^2-5q+2=0

q = 2 or q = 1 2

Substitution, got.

Q=2 when a3=4

q=1 2 at a3=-4