Question

若等比数列的前n项和是48,前2n项和是60,则前3n项的和是(    )

A.183              B.108              C.75                      D.63

Answer 1

解法一:S_n==48,

S_2n===S_n(1+qⁿ),

∴60=48(1+qⁿ),得qⁿ=.

∴S_3n==S_n(1+qⁿ+q²ⁿ)=63.

解法二:S_n、S_2n-S_n、S_3n-S_2n成等比数列,从而有S_n=48,S_2n-S_n=12,S_3n-S_2n=3,

∴S_3n=63.

解法三:令n=1,S₁=a₁=48,S_2n=S₂=a₁+a₂=48+a₂=60,

∴a₂=12,a₃=a₂·q=12×=3.

∴S_3n=a₁+a₂+a₃=48+12+3=63.

答案:D