We have,
a2 – a1 = 2a – a = a
a3 – a2 = 3a – 2a = a
a4 – a3 = 4a – 3a = a
i.e. ak+1 – ak is the same every time.
So, the given list of numbers forms an AP with the common difference d = a
Now, we have to find the next three terms.
We have a1 = a, a2 = 2a, a3 = 3a and a4 = 4a
Now, we will find a5, a6 and a7
So, a5 = 4a + a = 5a
a6 = 5a + a = 6a
and a7 = 6a + a = 7a
Hence, the next three terms are 5a, 6a and 7a