题目内容
等差数列{an}中,若a2+a3+a10+a11=32中,则a6+a7=( )
| A.9 | B.12 | C.15 | D.16 |
解法1:∵{an}为等差数列,设首项为a1,公差为d,
∴a2+a3+a10+a11=a1+d+a1+2d+a1+9d+a1+10d=4a1+22d=32,
∴2a1+11d=16,
∴a6+a7=a1+5d+a1+6d=2a1+11d=16;
解法2:∵a2+a11=a3+a10=a6+a7,a2+a3+a10+a11=32,
∴a6+a7=16,
故选D
∴a2+a3+a10+a11=a1+d+a1+2d+a1+9d+a1+10d=4a1+22d=32,
∴2a1+11d=16,
∴a6+a7=a1+5d+a1+6d=2a1+11d=16;
解法2:∵a2+a11=a3+a10=a6+a7,a2+a3+a10+a11=32,
∴a6+a7=16,
故选D
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