Any term of the first series can be written as 3+4(n-1) = 4n-1
Any term of the second series can be written as 1+5(m-1) = 5m-4
Here, m and n are natural numbers.
For commen term, 4n-1 = 5m-4
or 4n = 5m - 3...............(1)
Try putting m=1,2,3... and calculate n each time. 
eg. for m=1, 4n = 2, or n = 1/2. But 1/2 is not a natural number.
So try m=2, 4n = 10-3= 7. not a natural number
For m=3, 4n = 15-3 = 12. or n = 3.
So, condition (1) is first satisfied by n = 3. 
Obviously, since the LCM of 4 and 5 is 20, the next time it will be satisfied is when 5m-3 = 20+12. 
Basically, it will be satisfied whenever 5m-3 = 20k+12, where k is a WHOLE number. 
This gives n = (5m-3)/4 = 5k+3.
So, the common terms are 4n-1 = 4(5k+3)-1 = 20k+11
eg. the first sommon term is 11, when k=0. The second common term is 31, when k=1 and so on.
So the 10th common term will be 20(9)+11 = 191
 
A Shortcut
In all such questions, the nth common term is given by (LCM of common differences)*(n-1) + first common term
In this particular question, common differences of the 2 series are 4 and 5. Their LCM is 20. And the first common term is 11.
So the 10th common term is 20(10-1)+11 = 191.
This way you can also do questions where you have to find common terms of 3 or more series.
						

							
The first common term is 11.LCM of common difference=LCM(4,5)=20.consider an ap in which a=11,n=10and d=20. Use the formula to obtain the answer 191.