Given A.P.s are:

3, 7, 11, 15, 19, 23… 407   [common difference is 4]

2, 9, 16, 23… 709   [common difference is 7]

We can observe that the first common term is 23.

[6^th term in 1^st A.P. and 4^th term in 2^nd A.P. are the same.]

Let the number of common terms be 'n'.

They will be in A.P. with common difference = 28 [I.e. L.C.M(4,7)]

Now, the last common term can't be greater than 407.  [Last term of the first A.P]

$\therefore 23+\left(\mathrm{n}-1\right)28\le 407$

$\Rightarrow \mathrm{n}\le \frac{384}{28}+1$

$\Rightarrow \mathrm{n}\le 14.7$

$\Rightarrow \mathrm{n}=\mathbf{14}$