The first n terms of the series of equal differences are known and denoted as Sn, if a4 12, a8 4

Since it is an equal difference series, so a8-a4=4d, d is the tolerance, then d=-4, from a4=a1+3d, we can know a1=a4-3d=24, from sn=na1+n(n-1)d 2 to get sn=-2n 2+26n

That, I think it should be the maximum value, the tolerance is less than zero, so an is getting smaller and smaller, a7 is already 0, so s7 is the maximum value, to say the minimum value, then n keeps increasing, isn't it always decreasing.

Also, for the sn formula obtained above, it is a parabola with an opening pointing downward, so there is only a maximum value, there is no minimum value, since n has no range, so it should be the maximum value, right?

sn=-2(n-13 2) 2+169 2 (formula sn, find the maximum value); Therefore, when n=6 or 7, the sn is the largest, and the maximum value is 84

a1=24,a2=20,s3=16,a4=12,a5=8,a6=4,a7=0,a8=-4;

That's about it.

Summary. Hello, please send out the complete problem and let the sum of the first n terms of the difference series be sn, and a4+a8=a5+4, then s13 Hello, please send the complete problem to the sum of the first n terms of the difference series is sn, and a4+a8=a5+4, then s13 is good. And half more, right away.

Oooh. Can you answer another question?

OK. You can take a picture and send it out <>

Questions 3 and 4, thank you.

Okay, I'll look at the title.

Okay, thank you, teacher.

This is the third question.

a4+a8=2a6=0

So a6=0

s6=s5+a6

The object is subdued by s6=s5

ds5=s4+a5 in friend selection

A5 is not equal to the cover to carry 0, so B is not selected

Let the tolerance be d

d=（a8-a4）/4=2

a1=a4-3d=-18

SN minimum an variant.

an=a1+(n-1)d

a10=0, so s9=s10=-90 is the smallest.

It is a proportional series with a common ratio of 2 to the first term.

sn1=2(2^n-1)

The sum of the first n terms sought.

sn1+n（a1-d）

2(2^n-1)-20n

The minimum value is -90 n for 9 or 10

The sum of the first n terms of bn is 2 n + 1 -2 20n

an unequal difference series.

a8-a4=4d=8,d=2,a1=a4-3d=-18an, the general term is an=-20+2n

From the meaning of the question, bn=a2 (n-1)=-20+2*2 (n-1)=-20+2 n

Then the first n terms of bn and tn should be sought.

Let the tolerance be d

d=（a8-a4）/4=2

a1=a4-3d=-18

SN minimum an variant.

an=a1+(n-1)d

a10=0, so s9=s10=-90 is the smallest.

It is a proportional series with a common ratio of 2 to the first term.

sn1=2(2^n-1)

The sum of the first n terms sought.

sn1+n（a1-d）

2(2^n-1)-20n

d=2a1=-18

an=-18+(n-1)*2=2n-20

Obviously, the column is an increasing column.

Because sn=-18n+n(n-1)=n-19n=(n-19 2) -361 4 snmin=s9 or s10 is calculated to be a minimum of -90

a4+a6=-6.So a5=-3

a1=-11

4d=a5-a1=8

d=2sn=a1n+n（n-1）d/2

n²-12n

n-6)²-36

When n=6, it opens and closes openly. The hall has a minimum value of 36

a4+a6=2a5=-6 a5=-3 Through a1 and a5, you can calculate the formula of the general term d in the nullity, and then use it as a function to calculate the maximum value of the function.