In the series 3, 7, 11, 15, … and 2, 5, 8, … each continued to 100 terms, the number of terms that are identical is

Let the nth term of the first series = the mth term of the  second series.∴ 3+(n−1)×4=2+(m−1)×3or  4n=3m or n3=m4=k (say) ∴ n=3k and m=4kAs each series is continued to 100 terms, n=3k≤100 and m=4k≤100∴Possible values of k are 1, 2, 3, …, 25 and corresponding to each value of k we get one identical term. Hence there are 25 identical terms.