已知等差数列{an}中,a10=30,a20=50. (1)求通项公式; (2)若Sn=242,求项数n.
问题描述:
已知等差数列{an}中,a10=30,a20=50.
(1)求通项公式;
(2)若Sn=242,求项数n.
答
(1)a10=a1+9d=30,a20=a1+19d=50,解得 a1=12,d=2.∴an=a1 +(n-1)d=2n+10.…(6分)(2)∵Sn =na1+12n(n-1)d,∴242=12n+12n(n-1)•2,解得 n=11,或 n=-22 (舍去),故取n=11. …(12分)...