A1 20, An 54, Sn 999, find d and n

an=a1+(n-1)d=54

NA1+N(N-1)D=54N 1sn=NA1+N(N-1)D2=999 21-2.

n(n-1)d 2=54n-999 33 is substituted into 2.

na1+54n-999=999

20n+54n=1998

74n=1998

n=27 4

Equation 4 is substituted for a1 + (n-1) d = 54.

20+(27-1)d=54

d=17/13

Pairwise summing: a1+a99=1 (4+2 100)+1 (4 99+2 100) (ling2 100=t).

1/(4+t)+4/(t^2+4t)=(t+4)/(t^2+4t)=1/t=1/2^100

In the same way: a2+a98=a3+a97=·· 1 2 100 so the original formula = (1 2 100) 99 2 = 99 2 101

(1) Solution.

a1=20,an=a1+(n-1)*d=54,sn=a1+a2+a3+..an=a1+(a1+d)+(a1+2*d)+.a1+(n-1)*d)]=999

n*a1+[d+2d+..n-1)*d]=n*a1+(1+n-1)*(n-1)*d/2=n*a1+n*(n-1)*d/2

n*a1+n*(an-a1)/2

20n+34n/237n

So n=27, substituting an=a1+(n-1)*d=54, can be calculated as d=13 17

As long as you understand the solution of 1, you can easily solve the problem of 2, 3, 4...

2,3,4 figure it out by yourself, the difference series is not a problem for you...

(1) a1=20, an=a1+(n-1)*d=54 ,sn=n(a1+an) 2 i.e. 999=n 74 2, n=999 37=27, substituting n into d=17 13

2)d =1 3 ,n =37, sn=629, an=a1+(n-1)*d from an=a1+36 3 ,sn=n(a1+an) 2 629=37(a1+an) 2 an-a1=12,a1+an=34 please figure out the answer by yourself.

3 )a1=5 6 d=-1 6 sn=-5, an=5 6-(n-1) 6 from an=a1+(n-1)*d ,sn=n(a1+an) 2 -5=n(5 6+an) 2 solve the equation by yourself.

4) d=2 n=15 an=-10 to find a1 and sn can also be solved by the above formula.

n=10

You can also use formulas, and you can also ...... recursion