1.在等差数列｛an｝中,已知a2+a5+a8=9,a3a5a7=21,求a11.

题目

1.在等差数列｛an｝中,已知a2+a5+a8=9,a3a5a7=21,求a11.
2.若｛an｝为等差数列,公差为d1,｛bn｝为等差数列,公差为d2,证明：｛an+bn｝也为等差数列.
3.若｛an｝等差数列,a3+a6=4,a8+a11=9,求a10+a13.

答案

(1)an=a1+(n-1)da2+a5+a8=93a1+12d=9a1+4d=3 (1)a3a5a7=21(a1+2d)(a1+4d)(a1+6d)=21(3-2d)(3)(3+2d)=219-4d^2=7d^2=1/2d = ±√2/2d=√2/2 ,a1= (4-√2)/8 d=-√2/2 ,a1= (4+√2)/8 a11= a1+10d= (4-√2)/8 +(n-1)√...