how do we find the nth term for a sequence with two differences?

Question

can someone explain in simple words and steps the detailed process of how we can find the nth term of a sequence with two differences.

Answer 1

I assume you mean that the terms of the sequence are generated by adding two numbers alternatingly to the previous term.

So
n2 = n1 + a
n3 = n2 + b
n4 = n3 + a
n5 = n4 + b
...

The general formula (i = 1, 2, ...)
n(2i-1) = n1 + (a+b)*(i-1)
n(2i) = n1 + (a+b)*i - b

Check:
n(1) = n(2*1-1) = n1 + (a+b)*0 = n1
n(2) = n(2*1) = n1 + (a+b)*1 - b = n1 + a
n(3) = n(2*2-1) = n1 + (a+b)*1 = n1 + a + b
n(4) = n(2*2) = n1 + (a+b)*2 - b = n1 + a + b + a
etc.

Answer 2

can you give me an example of one and maybe i can help you.