It is known that the series of equal differences an satisfies a3 7, a1 a11 26, and the sum of the fi

The formula for the nth term of the equal difference series an=a1+d(n-1) (a1 is the first term, d is the tolerance, and n is the number of terms).

The first n terms of the equal difference series and the formula sn=(1 2)*(a1+an)*n (a1 is the first term, an is the last term, and n is the number of terms).

It can be seen from the question. a3=a1+2d=7

a1＋a11=a1+a1+10d=26

So a1+2d=7

a1+5d=13

So solve the system of binary linear equations about a1 and d.

a1=3 d=2

So an=3+2*(n-1)=2n+1

sn=(1/2)*(3+2n+1)*n =n(n+2)=n^2+2n

a3=7，a1+a11=2a6=26,a6=13d=(a6-a3)/3=2

a1=a3-2d=3

an=3+2(n-1)=2n+1

sn=(a1+an)n 2=(2n+4)n 2=n +2nThis is the conclusion I came to after meditating on it, if it can help you, I hope you will give me a (satisfaction).

If you can't ask, I'll do my best to help you out

It is not easy to answer the question, if you are dissatisfied, please understand

Let the tolerance be dd=(a6-a3) 6-3=2

a1=a3-2d=3

sn=na1+( n (n 1)))Ode to the tease of Ye Feng2 d3n+(n (n 1)refers to Xun 2) 2

n²+2n

a3+a5+a7=33 can be solved to obtain a5=11, and the tolerance d=(a3-a5) 2=2 can also be obtained by the first term a1=3

i.e. an=2n+1

sn=(an+a1)n/2=n(n+2)

A5+A7=2A6=26, we can see A6=13, and A6=A3+3D, find the tolerance D=2, and then an=A6+(n-6)D=13+(N-6)2=2N+1

Therefore, it is an odd sequence starting with 3, sn=(a1+an)*n 2=(2n+4)*n 2=2n+n 2.

Solution: a5+a7=2a6 So a6=26 2=13a6=a3+3d d=(a6-a3) 3=2a1=a3-2d=3

So: {an} is a series of equal differences with 3 as the prime minister and 2 as the tolerance.

an=a1+(n-1)d=3+2×(n-1)=2n+1sn=(a1+an)×n÷2=(3+2n+1)×n÷2=n(n+2)=n

Solution: Let the tolerance be d

a5+a7=2a6=26, (equal difference median).

So a6=13, and a6=a3+3d, d=2, an=a6+(n-6)d=13+(n-6)2=2n+1

again a1=3, so sn=(a1+an)*n 2=(2n+4)*n 2=2n+n 2.

1) Because a5 + a7 = 26

So a6=13, then the tolerance d=(13-7) (6-3)=2, so an=7+2(n-3)=2n+1

So a4=2)bn=1 carries 2n(2n+2)=1 4(1 n-1 (n+1)), so tn=1 pie 4(1-1 2+1 2-2 3+2 3-3 4+...1 n-1 (n+1))=1 4*n dust (n+1)=n 4(n+1).

Let the difference be d, then a5 = a3 + 2d, a7 = a3+ 4d, so there is:

a3 + 2d + a3 + 4d) = 26 substitute a3 = 7 into the above formula: d = 2 so a4 = a3 + d = 9

a1 = a3 - 2d = 3

So sn = na1 + n(n-1)d] 23n + n2 - n = n 2 + 2n

Solution 1Because it is an equal difference series, let the tolerance be d

So a5=a3+2d

a7=a3+4d

And because a5+a7=26

So a3+a3+2d+4d=26

Bring a3=7 in.

D=2 is obtained

Because a1+2d=a3

Bring a3=7 d=2 in.

A1=3 is obtained

So an=a1+(n-1)d

2n+12. sn=na1+(n-1)n/2 *d=n²+2n

1.Suppose the tolerance of the difference series is d

then a5 = a3 + 2d

a7=a3+4d

So a5+a7=2a3+6d=14+6d=26d=2, then a1=a3-2d=7-4=3

So the general formula is an=a1+(n-1)d

i.e. an=3+2(n-1).

3n+n(n-1)

n²+2n

The difference series satisfies 2a5=a3+a7=7+a7 and a5+a7=26 to obtain a5=11, a7=15, d=1 2(a5-a3)=1 2(11-7)=2

a1=a3-2d=7-4=3.

an=a1+(n-1)d=3+2(n-1)=2n+1sn=(a1+an)n 2=(3+2n+1)n 2=n 2+2nn 2 represents the square of n.

In the case of the difference series, a3 = a1 + 2d = 7, a5 = a7 = 2a1 + 10d = 26, the sister equation system gets d = 2, a1 = 3, so an=1 + 2n

sn=n(a1+an)/2=n^2+2n

Arithmetic progression.

So a3+a7=a4+a6=0

a3a7=-16

By the Vedic theorem.

a3 and a7 are the two roots of the equation x -16 = 0.

x=4 if a3=4, a7=-4

then a7-a3==4d=-8

d=-2a1=a3-2d=8

an=a1+(n-1)d=-2n+10

sn=(a1+an)n/2=n²+9n

If a3=-4, a7=4

then a7-a3==4d=-8

d=2a1=a3-2d=-8

an=a1+(n-1)d=2n-10

sn=(a1+an)n/2=-n²-9n

So sn=n +9n or sn=-n -9n

Let the tolerance be d

a3a7=-16

a1+2d)(a1+6d)=-16...1a4+a6=0

a1+3d+a1+5d=0...2 Simultaneous 1,2 solutions get a1=-8, d=2 or a1=8, d=-2, so an=2n-10 or an=10-2n

sn=n(a1+an)/2

sn=n -9n or sn=-n +9n