The higher difference number series problem is an ingenious method

1.Because a1>0, s9>0 s10<0, this series is a decreasing series.

s9＝9*（a1+a9)/2>0

s10=10*(a1+a10)/2<0

So a1+a9>0

a1+a10<0

a5+a5>0

a5+a6<0

So a5>0 a6<0

When n 5, the sn is maximum.

2.It can be done using the above methods.

a3+a8>0

So a5+a6>0, a6>0

s9<0

a1+a9）*9/2<0

a1+a9）=2a5<0

So a5<0

When n=5, s5 has a minimum value.

1.Obviously, the tolerance is less than 0, s9>0, (a1+a9)*9 2>0, (a1+a9)=2a5>0, a5>0

s10<0,，（a1+a10）*9/2<0,（a1+a9）=a5+a6<0,a6<0

Finding the SN Minimum? The SN will always be smaller because the tolerance is negative.

Shouldn't I ask if there is a maximum value of sn, then when n=5, there is a maximum value, because all positive numbers are added, and then there will be negative numbers.

0,a5+a6>0,a6>0 (as can be seen below).

s9<0，（a1+a9）*9/2<0,（a1+a9）=2a5<0,a5<0

Then when n=5, s5 has a minimum value, because all negative numbers are added, and then added there will be positive numbers.

Question 1 is incorrect, and the maximum value of sn should be found when n=(9+1) 2=5.

From 6a1+(n+12)d=24, a1=(24-(n+12)d) 6 (24-(n+12)d) 6+5d=4-(n+12)d, 6+5d=4+(5-(n+12) 6)d, is the fixed value.

d is a variable. 5-（n+12）/6=0

n=18

Compare 6A1+(N+12)D=24, S11=11A1+55D=11(A1+5D), if S11 is a fixed value, then A1+5D is a fixed value. 6 must be mentioned in 6a1+(n+12)d=24, so divide by 6 to get 6(a1+(n+12)d 6)=6*4.

Mastering the flexible use of formulas is one of the required questions in the college entrance examination.

1.a3=a1+2d=1+2ddA5=a1+4d=1+4db3=b1q =q b5=b1q 4=q 4.

1+2d+q^4=21

1+4d+q^2=13

The solution is d=2, q=2 (q=-2 rounded).

So an=a1+(n-1)d=1+2(n-1)=2n-1bn=b1q (n-1)=2 (n-1).

2) an/bn=(2n-1)/2^（n-1)

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Because a(n+1)=5an (5+an).

So 1 a(n+1)=(5+an) 5an, so 1 a(n+1)-1 an=(5+an) 5an -1 an=1 5=d

So into a series of equal differences, so 1 an=1 a1+nd=1+n 5, so an=1 (1+n 5).

- -Is your an+1 a whole or separate?

Equal difference series an=a1+(n-1)d

Because a4+a6=a1+3d+a1+5d=2(a1+4d)=0, the dandan, a1+4d=0

Because a3*a7=(a1+2d)(a1+6d)=(a1+4d-2d)(a1+4d+2d)=(2d)*2d=-4d 2=-16

The difference is measured by d= 2, a1=-8 or 8

i.e. an=8+(-2)(n-1).

or an=-8+2(n-1).

So sn=[-8+(-8+2(n-1))]n 2=n 2-9n or analogue sn=[8+(8+(-2)(n-1))]n 2=9n-n 2

By a4 + a6 = 2 a5 = 0, the first gear is therefore a5 = 0

and a3*a7 (a5-2d) (a5+2d) = -4d 2 = -16, d = 2 or -2

1) When d=2, a1=-8, an=a5+(n-5)d=2n-10sn=(a1+an)n 2=n(n-9).

2) When d=-2, a1=8, an=a5+(n-5)d=-2n+10sn=(a1+an)n trillion virtual 2=-n(n-9).

Hello, look**, ask for tolerance first. Then calculate that the value after that term is less than . Then count the sum.

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As a result, ** can't be sent.

Because. a(n+1)=5an/(5+an)

So. 1 a(n+1)=(5+an) 5an, so 1 a(n+1)-1 an=(5+an) 5an1 an=1 5=d

So into a series of equal difference sales.

The search year is 1 an=1 a1+nd=1+n 5, so an=1 (1+n 5).

Because (an) is a series of equal differences, s10, s20-s10, s30-s20....S100-S90, S110-S100 are also equal difference series, and tolerance D is 100 times that of AN. (i.e. d=100d).

s100=s100-s90+..s20-s10+s10=s10+s10+d+s10+2d+..s10+9d=10*s10+45d

d=(s100-10*s10)/45=(10-10*100)/45=-22

s110=s110-s100+s100-s90+..s20-s10+s10=s10+s10+d+s10+2d+..s10+9d+s10+10d=11s10+55d=11*100-55*22=-110

（1）a6+a9+a12+a15=2(a1+a20)=34a1+a20=17

s20=20(a1+a20)/2=10*17=170(2)d=(910-310)/100=6

s30-s20)-910=100d=600∵a11=a1+10d

s20-s10=10d*10

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