首項(xiàng)為的等差數(shù)列,從第項(xiàng)開(kāi)始為正,則公差的取值范圍是 ( )C A. B. C. D. 查看更多

 

題目列表(包括答案和解析)

(2009•長(zhǎng)寧區(qū)一模)無(wú)窮等比數(shù)列{an}中,
lim
n→∞
(a1+a2+…+an)=
1
2
,則首項(xiàng)a1的取值范圍(  )

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已知等差數(shù)列{an}的首項(xiàng)a1>0,公差d>0,前n項(xiàng)和為Sn,設(shè)m,n,p∈N*,且m+n=2p
(1)求證:Sn+Sm≥2Sp
(2)求證:Sn•Sm≤(Sp2;
(3)若S1005=1,求證:
2009
n=1
1
Sn
≥2009

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(2009•閘北區(qū)一模)記數(shù)列{an}的前n項(xiàng)和為Sn,所有奇數(shù)項(xiàng)之和為S′,所有偶數(shù)項(xiàng)之和為S″.
(1)若{an}是等差數(shù)列,項(xiàng)數(shù)n為偶數(shù),首項(xiàng)a1=1,公差d=
3
2
,且S″-S′=15,求Sn
(2)若{an}是等差數(shù)列,首項(xiàng)a1>0,公差d∈N*,且S′=36,S″=27,請(qǐng)寫(xiě)出所有滿(mǎn)足條件的數(shù)列;
(3)若數(shù)列{an}的首項(xiàng)a1=1,滿(mǎn)足2tSn+1-3(t-1)Sn=2t(n∈N*),其中實(shí)常數(shù)t∈(
3
5
,3)
,且S-S=
5
2
,請(qǐng)寫(xiě)出滿(mǎn)足上述條件常數(shù)t的兩個(gè)不同的值和它們所對(duì)應(yīng)的數(shù)列.

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(2009•朝陽(yáng)區(qū)二模)設(shè)數(shù)列{an}的首項(xiàng)a1=1,前n項(xiàng)和為Sn,且點(diǎn)(Sn-1,Sn)(n∈N*,n≥2)在直線(xiàn)(2t+3)x-3ty+3t=0(t為與n無(wú)關(guān)的正實(shí)數(shù))上.
(Ⅰ) 求證:數(shù)列{an}是等比數(shù)列;
(Ⅱ) 記數(shù)列{an}的公比為f(t),數(shù)列{bn}滿(mǎn)足b1=1,bn=f(
1
bn-1
)
(n∈N*,n≥2).
設(shè)cn=b2n-1b2n-b2nb2n+1,求數(shù)列{cn}的前n項(xiàng)和Tn;
(Ⅲ) 在(Ⅱ)的條件下,設(shè)dn=(1+
1
3bn-1
)n
(n∈N*),證明dn<dn+1

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(2009•上海模擬)已知無(wú)窮等比數(shù)列{an}的前n項(xiàng)和為Sn,各項(xiàng)的和為S,且
lim
n→∞
(Sn-2S)=1
,則其首項(xiàng)a1的取值范圍是(  )

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