In the formula an a1 n 1 d, n is a natural number, known a2 4, a5 14, find a10

The substitution value is: 4=a1+(2-1)d

14=a1+(5-1)d

Solving the system of equations yields a1=10, d=-6

So a10 = 10 + (10-1) * - 6 = -44

It can be seen that an is a series of equal differences.

So the work is poor. d=(a5-a2)/3=-6

So a10=a2+8d=4-6*8=44

a2=a1+d=4.

a5=a1+4d=-14.

A1=10 and d=-6 are obtained from the first and second formulas

Substitute a10 = a1 + 9d = 10 + 9 * (-6) = -44

Solution: a2=a1+(2-1)d

a5=a1+(5-1)d

Solve binary linear equations.

A1 and D are obtained and then substituted into the original equation an=a1+(n-1)d to obtain a10.

Kid, I don't need to teach you later.

Answer: a1=1

a(n+1)=an*(-1)^n

a(n+1)/an=(-1)^n

When the balance n is odd: a(n+1) an=-1 and when n is even: a(n+1) an=1

So: the odd-numbered term is equal to the preceding even-numbered term, and the even-numbered term is the opposite of the preceding odd-numbered term.

So the series are arranged according to the following law:

That is, 1, the vertical Zheng Ming cong cavity cycle appears.

2007 4=501 remainder 3

So: 2007th Term-1

So: a2007=-1

There is a problem with the question, such as a1=-1, a2=-2, a3=3, n=3a1+a2+a3) n=0=0

Let g(x)=a1+a2+.an=k,f(x)=a1a2...ANF(x) is the maximum Bifocal difference, DF=MDG

a2a3a4...an=m

a1a3a4...an=m

a1a2a4...an=m

a1a2a3...a(n-1)=m

a1+a2+..an=k

Argument Wood a1=a2=.an=k/n

i.e. a1a2....an

Answer: The second question: divide by 2 n when the two sides negotiate the same round to get an 2 n-a(n-1) 2 (n-1) = 1 is the first term is a1 2 0.

1 is a series of equal differences for tolerance.

i.e. an 2 n=n

Therefore an=n*2 n

The third question can be solved directly by dislocation subtraction.

I've written it too many times.

Practice it yourself.

When in doubt.

When n 2, roll early.

an=2a(n-1)+2^n

Divide by 2 n on both sides at the same time

An 2 n=a(n-1) 2 (n-1)+1an 2 n-a(n-1) 2 (n-1)=1 is. Arithmetic progression.

The tolerance of the crack is 1

The first term a1 2 0=2

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

an=(n+1)*2^n

Hello. First, replace a2 and a5.

Count the muffled opening of the oak out of d = 3

a1=-4, so an=-4+3 (n+1) substituting n=10.

You can ask for the a10=29 thank you.

Hello. First of all, substitute Jane A2 and A5.

Calculate d=3a1=-4

The repentant brigade is an=-4+3(n+1).

Substitute n=10.

You can find the a10=29 buried before the block, thank you.

Let bn=a(n+1)-an

2a(n+2)=an+a(n+1)

2[a(n+2)-a(n+1)]=an-a(n+1)=-[a(n+1)-an]

bn=a(n+1)-an, 2b(n+1)=-bn, i.e. b(n+1) bn=-1 2

It's a proportional series.

b1=a2-a1=2-1=1, which is the first proportional series with 1 and the common ratio of -1 2.

bn=1*(-1/2)^(n-1)

a(n+1)-an=(-1/2)^(n-1)

an-a(n-1)=(-1/2)^(n-2),a(n-1)-a(n-1)=(-1/2)^(n-3)

a2-a1=(-1/2)^0

The above superimposes an-a1=(-1 2) 0+......1/2)^(n-3)+(1/2)^(n-2)

1-(-1/2)^(n-1)]/(1+1/2)=(2/3)[1-(-1/2)^(n-1)]

an=a1+(2/3)[1-(-1/2)^(n-1)]=5/3-(2/3)*(1/2)^(n-1)=5/3+(1/3)*(1/2)^(n-2)

1999 Master Meng Meng, you are so cheap, don't give an answer.