Question

在等差数列中，记公差为d，前n项的和为sn，已知a6-a3=1，4s6=11s3，求通项an

Answer 1

6(a1+a6)/2=s6
3(a1+a3)/2=s3
because 4s6=11s3
24(a1+a6)=33(a1+a3)
24a1-33a1+24a6-33a3=o
because a6-a3=1
thus -9a1-9a3+24=0
24-18a2=0
solve a2, 3a2=s3, s6 can be solved too, and then u will see how to do it

what a boring day~~~~
i wanna back to china!

Answer 2

a6-a3=3d=1
所以d=1/3
所以Sn=na1+n(n-1)/(2*3)=na1+n(n-1)/6
所以S6=6a1+6*5/6=6a1+5
S3=3a1+3*2/6=3a1+1
又因为4S6=11S3
所以11*(3a1+1)=4*(6a1+5)
解得a1=1
所以an=1+(n-1)/3=n/3+2/3

Answer 3

在等差数列中，记公差为d，前n项的和为sn，已知a₆-a₃=1，4S₆=11S₃，求通项an
解:a₆-a₃=(a₃+3d)-a₃=3d=1,故d=1/3,
4[6a₁+6×5×(1/3)/2]=11[3a₁+3×2×(1/3)/2]
即24a₁+20=33a₁+11
故9a₁=9, ∴a₁=1
故通项a‹n›=1+(n-1)(1/3)=(n+2)/3