find the number of common terms in the sequence 3,7,11,......,407 and 2,9,16,......,709

let the number of terms in two AP's are m and n respectively.
$$\begin{gathered} 407 is the mth term of the AP 3,7,11,..... \\ 407=3+(m-1)*(7-3) \\ 407-3=4(m-1) \\ 404=4(m-1) \\ m-1=\frac{404}{4}=101 \\ m=1+101=102 \end{gathered}$$
$$\begin{gathered} 709 is the nth term of the AP 2,9,16,..... \\ 709=2+(n-1)*(9-2) \\ 709-2=7(n-1) \\ 7(n-1)=707 \\ n-1=\frac{707}{7}=101 \\ n=1+101=102 \end{gathered}$$
therefore the number of terms in each AP are 102.
let the pth term of first AP is equal to the qth term of 2nd AP
$$\begin{gathered} 3+(p-1)*4=2+(q-1)*7 \\ 4p-1=7q-5 \\ 4p+4=7q \\ 4(p+1)=7q \\ \frac{p+1}{7}=\frac{q}{4}=k \\ \Rightarrow p=7k-1 and q=4k \end{gathered}$$
now p and q must be less than 102.[the total number of terms in each AP are 102]
$$\begin{gathered} 7k-1\le 102 \& 4k\le 102 \\ 7k\le 103 k\le \frac{102}{4} \\ k\le \frac{103}{7} k\le 25\frac{1}{2} \\ k\le 14\frac{5}{7} \end{gathered}$$
thus k=1,2,3,.....14
thus the number of identical terms are 14.

hope this helps you