Question

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已知等差数列｛an｝的前n项和为Sn，若a3＋a4＋a5＝12，则S7的值为

Answer 1

等差则a3+a5=2a4
所以3a4=12
a4=4

a1+a7=2a4=8
所以S7=(a1+a7)*7/2=28

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Answer 2

解由a3＋a4＋a5＝12，
即3a4=12
即a4=4
S7=1/2*7(a1+a7)
=1/2*7*2a4
=7a4
=7*4
=28

Answer 3

a3＋a4＋a5＝12
3a4 = 12
a4 = 4

S7 = 7a4 = 7 *4 = 28

Answer 4

因为是等差数列，所以a3＋a5=2a4，若a3＋a4＋a5＝12，则a4=4.
S7=a1＋a2＋a3+a4＋a5＋a6+a7=（a1＋a7）+（a2＋a6）+（a3＋a5）+a4=2a4+2a4+2a4+a4=7a4=7×4=28

Answer 5

因为是等差数列，所以3a4=12 所以a4=4,
因此2a4=8,
而a1+a7=a2+a6=a3+a5=2a4
得a1+a2+a3+a4+a5+a6+a7=7a4=28

Answer 6

a3＋a4＋a5＝3a1+9d=12, a1+3d=4;
S7=7a1+21d=7(a1+3d)=28