Question

设sn为等差数列an的前n项和 ，已知S3=a7,a8-2a3=3

Answer 1

回答的较为繁琐但比较详细，每一步很清晰的。
设an=a1+(n-1)d
已知s3=a7
a1+6d=a1+(7-1)d=a7=s3=a1+a2+a3=a1+[a1+(2-1)d]+[a1+(3-1)d]=3a1+3d
即：a1+6d=3a1+3d
→2a1=3d

已知:a8-2a3=3
→3=a8-2a3=[a1+(8-1)d]-2[a1+(3-1)d]=3d-a1=2a1-a1=a1
即：a1=3， d=2
→an=3+(n-1)×2=2n+1
还有哪步没懂请追问
若有帮助到您，请采纳，谢谢。

Answer 2

设an=a1+(n-1)d
a1+6d=a1+(7-1)d=a7=s3=a1+a2+a3=a1+[a1+(2-1)d]+[a1+(3-1)d]=3a1+3d→2a1=3d
3=a8-2a3=[a1+(8-1)d]-2[a1+(3-1)d]=3d-a1=2a1-a1=a1→d=2
an=3+(n-1)×2=2n+1