Question

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

Answer 1

(1)
an=a1+(n-1)d
a2+a5+a8=9
3a1+12d=9
a1+4d=3 (1)
a3a5a7=21
(a1+2d)(a1+4d)(a1+6d)=21
(3-2d)(3)(3+2d)=21
9-4d^2=7
d^2=1/2
d = ±√2/2
d=√2/2 ,a1= (4-√2)/8
d=-√2/2 ,a1= (4+√2)/8
a11= a1+10d
= (4-√2)/8 +(n-1)√2/2 or (4+√2)/8 -(n-1)√2/2
=(√2/2)n +(4-5√2)/8 or (-√2/2)n +(4+5√2)/8
(2)
an =a1+(n-1)d1
bn= b1+(n-1)d2
=an +bn
= (a1+b1)+ (n-1)(d1+d2)
{}是等差数列,d = a1+d2,c1=a1+b1
(3)
an = a1+(n-1)d
a3+a6=4
2a1+7d=4 (1)
a8+a11=9
2a1+17d=9 (2)
(2)-(1)
10d=5
d= 1/2
a1=1/4
a10+a13
=2a1+21d
=1/2 + 21/2
=11,10,1.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.