已知等差数列{an}中,a1=1,a3=-3 已知等差数列{an}中，a1=1，a3=-3.
\u5df2\u77e5\u5728\u7b49\u5dee\u6570\u5217an\u4e2d,a3=-3,a5+a10=30\u5047\u8bbe\u516c\u5dee\u662fd\uff0ca5+a10=30\u5c31\u7b49\u4e8e(a3+2d)+(a3+7d)=30,a3=-3,\u53ef\u5f97d=4\uff0c
an=-11+\uff08n-1\uff094
\u5c06\u6570\u503c\u5e26\u5165\u5f97540=-11n+n\uff08n-1\uff094/2
\u5316\u7b80\u5f972\uff08n2\uff09-13n-540=0
\u89e3\u5f97n=20
a3-a1=2d=-3-1=-4
\u2234d=-2
\u2235\ufe5ban\ufe5c\u662f\u7b49\u5dee\u6570\u5217
\u2234an=a1+(n-1)d=1-2(n-1)=-2n+3
\u660e\u6559\u4e3a\u60a8\u89e3\u7b54\uff0c
\u5982\u82e5\u6ee1\u610f,\u8bf7\u70b9\u51fb[\u91c7\u7eb3\u4e3a\u6ee1\u610f\u56de\u7b54];\u5982\u82e5\u60a8\u6709\u4e0d\u6ee1\u610f\u4e4b\u5904,\u8bf7\u6307\u51fa,\u6211\u4e00\u5b9a\u6539\u6b63!
\u5e0c\u671b\u8fd8\u60a8\u4e00\u4e2a\u6b63\u786e\u7b54\u590d!
\u795d\u60a8\u5b66\u4e1a\u8fdb\u6b65!
a3=a1+2d=-3
因a1=1 所以d=-2
(1) 通项公式an=a1+(n-1)d=1-2(n-1)=3-2n
(2) 前k项和Sk=(a1+ak)*k/2
=(1+3-2k)*k/2=-35
k^2-2k-35=0
(k-7)(k+5)=0
k=-5(舍去)
k=7
即为所求
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已知数列《an》中，a1=1，an，an+1，2a1成等差数列。《1》求a2,a3,a4;
(1)an=3-2n
(2)5
(1)an=a1+(n-1)d
a3=1+2*d=-3
d=-2
an=-2n+3
(2)sk=a1k+k(k-1)d/2=-35
k=7
绛旓細瑙ｅ緱d=-2鎴杁=0 褰揹=0鏃禸1=a2=1,bn=1 褰揹=-2鏃讹紝b1=a2=-1,b2=a3=-3,q=3锛岄氶」bn=-3^(n-1)
绛旓細鏍规嵁绛夊樊鏁板垪鍏紡鍙緱锛歛7=a1+6d锛屽甫鍏1=-3锛宎7=21锛屾眰寰梔=4
绛旓細閫氶」鍏紡锛an=2+锛坣-1锛壝1=n+1锛沘3=3+1=4,a5=5+1=6锛a10=10+1=11
绛旓細鍥犱负a1銆乤2銆乤5鎴愮瓑姣鏁板垪 鎵浠 a2² = a1 脳 a5 鍗 (a1 + d)² = a1 脳 (a1 + 4d)鎵浠 (1 + d) ² = 1脳(1 + 4d)d² + 2d + 1 = 4d + 1 d(d - 2) = 0 d = 2 (d = 0 鑸嶅幓)鎵浠 = 2 ...
绛旓細d=(a3-a1)/2=(10-2)/2=4 鎵浠 an=a1+(n-1)d=2+4(n-1)=4n-2 a8=4脳8-2=30 s8=(a1+a8)脳8梅2=锛2+30锛壝4=128
绛旓細鈭磏=(an-a2)/d+2 =(103-3)/5+2 =22 鍙堚埖a1=a2-d=-2 鈭碨n=(a1+an)n/2 =(-2+103)x22/2 =1111 3銆佲埖S31=31(a1+a31)/2 =31*(a1+40)/2 =2170 鍗31(a1+40)=4340 瑙ｅ緱锛歛1=100 鈭村叕宸甦=(a31-a1)/30 =(40-100)/30 =-2 4銆佲埖S41=31(a1+a41)/2 =41*(200+...
绛旓細a1=2,a4=a1+3d=2+3d=8 鍗砫=2 a2=a1+d=4 a3=a1+2d=6 鎵浠4=a1+a2+a3+a4=20
绛旓細璁鹃椤逛负a1,鍏樊涓篸 鎵浠 a6=a1+锛6-1锛塪=1 鎶奱1浠ｅ叆涓婂紡瑙ｅ緱 d=1/2
绛旓細a1=2,a3=10.a3=a1+2d=2+2d=10 d=4,a8=a1+7d=2+7脳4=30
绛旓細an = a1 + (n-1) d = 1+ (n-1)d Sn = (1 + 1+(n-1)d) * n /2 = (2+(n-1)d) n/2 a2n = a1 + (2n-1)d = 1 +(2n-1)d S2n = (1 + 1+(2n-1)d) * (2n)/2 =(2+(2n-1)d) *(2n) /2 S2n/ Sn = (2+(2n-1)d)*2 / (2+(n-1)d)S2/S1...
