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.