Question

求等差数列{an}：5、8、11、…、302与等差数列{bn}:3、7、11…、299的公共项构成的数列{cn}的通项公

Answer 1

an = 2+3n

bm = -1+4m m=1,2,..75

an =bm = 2+3n= -1+4m 4m=3(n+1)

∵ （3,4）=1 ,∴ 3|m m=3k (k=1,2,...,25)

4m=4*3k=3(n+1) n=4k-1 (k=1,2,...,25)

cn = 2 + 3*(4k-1) = 12k-1 (k=1,2,...,25)