3,7,11,…,139
通项3+4(n-1)=4n-1
4n-1<=139,n<=35
2,9,16,…,142
通项2+7(m-1)=7m-5
7m-5<=142,m<=21
4n-1=7m-5
7m=4n+4
所以m能被4整除
1<=m=4k<=21
所以k=1,2,3,4,5
则28k=4n+4
7k=n+1
1<=n=7k-1<=35
k=1,2,3,4,5
所以k=1,2,3,4,5
所以公共项有5项