§6.3 等比數(shù)列專題檢測(cè)1.(2019北京朝陽(yáng)二模,5)已知等差數(shù)列{an}的首項(xiàng)為a1,公差d≠0,則“a1,a3,a9成等比數(shù)列”是“a1=d”的( )A.充分而不必要條件 B.必要而不充分條件C.充要條件 D.既不充分也不必要條件答案 C 由a1,a3,a9成等比數(shù)列,得=a1a9,從而(a1+2d)2=a1(a1+8d),又d≠0,所以a1=d;若a1=d,則a3=3a1,a9=9a1,從而有=a1a9,所以a1,a3,a9成等比數(shù)列.故“a1,a3,a9成等比數(shù)列”是“a1=d”的充要條件,故選C.2.(2018河南新鄉(xiāng)二模,6)在公比為q的正項(xiàng)等比數(shù)列{an}中,a4=4,則當(dāng)2a2+a6取得最小值時(shí),log2q=( )A. B.- C. D.-答案 A 由題意知a2>0,a6>0,所以2a2+a6≥2=2=8,當(dāng)且僅當(dāng)q4=2時(shí)取等號(hào),所以log2q=log2=,選A.3.(20195·3原創(chuàng)沖刺卷三,5)已知數(shù)列{an}為正項(xiàng)等比數(shù)列,a2=,a3=2a1,則a1a2+a2a3+…+anan+1=( )A.(2+)[1-] B.(2+)[-1]C.(2n-1) D.(1-2n)答案 C 由{an}為正項(xiàng)等比數(shù)列,且a2=,a3=2a1,可得a1=1,公比q=,所以數(shù)列{anan+1}是以為首項(xiàng),2為公比的等比數(shù)列,則a1a2+a2a3+…+anan+1==(2n-1).故選C.4.(2018福建廈門模擬,8)設(shè)等比數(shù)列{an}的前n項(xiàng)和為Sn,若Sn=2n+1+λ,則λ=( )
A.-2 B.-1 C.1 D.2答案 A 解法一:依題意,a1=S1=4+λ,a2=S2-S1=4,a3=S3-S2=8,因?yàn)閧an}是等比數(shù)列,所以=a1·a3,所以8(4+λ)=42,解得λ=-2.故選A.解法二:Sn=2n+1+λ=2×2n+λ,易知q≠1,因?yàn)閧an}是等比數(shù)列,所以Sn=-qn,據(jù)此可得λ=-2.故選A.5.(20195·3原創(chuàng)沖刺卷八,5)已知等比數(shù)列{an}滿足a1+a2=12,a1-a3=6,則當(dāng)a1·a2·…·an取到最大值時(shí),n的值為( )A.3 B.4 C.3或4 D.5答案 C 設(shè)等比數(shù)列{an}的公比為q,由a1+a2=12,a1-a3=6,可得解得∴an=8×=(n∈N*),∴a1·a2·…·an==,令f(n)=n(n-7)=(n2-7n)=-,當(dāng)n=3或n=4時(shí),f(n)有最小值,且f(n)min=-6,故當(dāng)n=3或4時(shí),a1·a2·…·an取得最大值,故選C.6.已知各項(xiàng)均為正數(shù)的等比數(shù)列{an},a1>1,0
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