In a series of equal differences with a tolerance of non zero, a3, a6, and a7 are proportional serie

Let the tolerance be d a3, a6, a7 into an equal proportional series.

then a6 = a3 a7, i.e. (a1+5d) = (a1+2d) (a1+6d).

A1 +10A1D+25D =A1 +8A1D+12D 2A1D+13D =0 2A1+13D=0 A1=13D 2, then A4 A6=(A1+3D) (A1+5D)=(13D 2+3D) (13D 2+5D).

Let the tolerance be d, then there is a5=a1+4d, a7=a1+6d, because it is known that a1, a5, a7 are equal to the number of limb ratios, so (a1+4d) 2=a1(a1+6d).

This gives a1=-8d

Thus the dry hole a4=a1+3d=-5d, a3=a1+2d=-6d, a6=a1+5d=-3d, so (a1+2a4) (a3+a6)=(8d-10d) (6d-3d)=2

Let the first item of the pin inspection be A1 and the tolerance is D>0, then.

a3*a7＝－12

a4+a6 a3 a7 4, so a3 6, a7 2(d> mindless 0) so a7 loses high a3 4d 8

So d 2 so a1 10

s20=-200+20*19/2*2=180

Summary. Kiss: The second question is to find the sum of the leading terms of the nth power of six and the sum of the leading terms of 3n, and then subtract them to get the sum of the first n terms of bn.

In a series of equal differences with a tolerance of not 0, a3+a4=7, a1,a4,a16 are proportional series.

Hello quietly cronyism ( Please open and ask if you can take a picture of the original question, which is more conducive to shouting and solving the problem

Good. Good.

This is the first question to ask Ha Qin.

Well. This is the second question.

Can the first question be answered?

Kiss: The first question is that we can find the coarse relationship between the boring number modulus d and a1 according to the equal ratio, and then we can find out how much a1 is equal to according to a 3 + a 4 = 7.

Kiss: The second question is to ask for the sum of the frontier terms of the nth power of the six and the frontier terms of 3n, and then subtract them after the hand regrets to get the sum of the first n terms of Bu Shu Zheng bn.

Kiss: Didn't the first question give you <>?

Kiss: When doing the problem of equal difference series or equal relative let tremor ratio series, we need to know what formulas are required to sum the equal difference slip ridge and equal ratio, and we need to remember what methods we can use to sum the general term formulas and what methods we have. For example, the sum of split terms, the dislocation subtraction, and so on.

Good. Thank you.

It's okay.

Known: a4=a1+(4-1)d=6,d=(6-a1) 3.

a2=a1+（2-1）d

a1+（6-a1）/3

3a1+6-a1）/3

2a1+6）/3

a3=a1+（3-1）d

a1+2（6-a1）/3

3a1+12-2a1）/3

a1+12）/3

It is also known that a2, the deficit a4, and a3 are proportional series, a4 a2=a3 a4, a4 =a2xa3=6 ,2a1+6) 3x(a1+12) 3=36,2a1 +24a1+6a1+72=36x9=324,a1 +15a1+36=162,a1 +15a1-126=0,a1+21)(a1-6)=0

Then: a1'=-21;al〞=6。

1) Replace a1' with d to get:

d'=(6-a1) 3=(6+21) 3=9s6'=(a1+a6)n

x6/2（-21+24）x3

2) Replace a1 with d to get

d = (6-a1 ) finch change 3=(6-6) 3=0, because the tolerance in the question is not zero, therefore, a1"=6 is meaningless.

Let the difference in the series of equal differences be d, then a1=5-2d; a2=5-d;a5 = 5 + 2d and because a1, a2, a5 are proportional sequences, so a2 a1 = a5 a2, substitute the above formula to obtain.

d1=2, d2=0 (should be known {an} is a series of equal differences with a tolerance of non-zero, so rounded) The general formula is an=a1+(n-1)d=2n-1.

The sequence {2 an} is the sum of the first n terms of the proportional sequence {2 an} with the first term 2 and the common ratio of 4 and sn = 2[(4 n)-1] 3

(a3-d)^2=(a3-2d)(a3+2d)

Think for yourself based on this. I went to sleep and was too sleepy.

Solution: (1) Let the tolerance be d, then a2 2-a5 2=a4 2-a3 2 get a1+5d=0, and s7=7 get a1+3d=1 to get a1=-5, d=2

The general term an=2n-7, the former term and sn=n 2-6n(2)ama(m+1) a(m+2)=(2m-7)(2m-5) (2m-3)=2m-9+8 (2m-3).

For this number to be an integer, 2m-3 must be an odd divisor of 8, so 2m-3=-1 or 1

When 2m-3=-1, m=-1 is not a positive integer, and when 2m-3=1, m=2 satisfies the topic.

In summary, m=2

s7=7(a1+a7)/2=7a4=7

a4=1 and by the conditional formula.

a2^2+a3^2=a4^2+a5^2

Simplification, i.e., a2 2-a5 2=a4 2-a3 2 gives (a2+a5)(a2-a5)=(a4+a3)(a4-a3), and a2+a5=a4+a3 is true.

When a2+a5≠0

We get a2+a3=a4+a5

That is, d=0 does not match the meaning of the title, so a2+a5=a4+a3=0

and a4 = 1 a3 = -1

an=-5+(n-1)*2

2n-7sn=-5n+n(n-1)

n^2-6n

Let the formula = 2k-7 form and 2k-7

That is, the formula is an odd number.

And the original formula = (2m-3)-6+[8 (2m-3)], so 2m-3 is a factor of 8.

m=2 is obtained

(1):an=a1+(n-1)d,d≠0

From a2 +a3 =a4 +a5 we know that 2a1+5d=0 and because: s7=7, a1+3d=1

→a1=-5,d=2

an=2n-7,sn=n(a1+an) 2=n -6n(2) because amam+1 am+2 = (am+2 -4)(am+2 -2) (am+2 )=am+2 -6+8 am+2 are the terms in the series. Therefore, 8 am+2 is an integer, and (1) we know that am+2 is an odd number (only 1) am+2 =2m-3= 1, i.e. m=1,2

Only m=1,2 fits the topic

a3*a7=-12

a3+a7=a4+a6=-4

Solve a3=2, a7=-6

or a3 = -6 and a7 = 2

a7=a3+4d

d=(a7-a3)/4

So d = -2 or d = 2

Since the tolerance is positive, d=2

a3=-6,a7=2

a1=a3-2d

a1=-10

sn=na1+n(n-1)d/2

s20=20*(-10)+20*19*2/2=-200+380=180