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Please notice that an arithmetic sequence looks like this:
a_1,a_1+d,a_2+d,...,a_{n-1}+d \\ a_1+d,a_1+2d,a_1+3d,...,a_n(n-1)d

Look at the terms:
1st term
2nd term = 1st term + d
3rd term = 2nd term + d = 1st term + 2d
4th term = 3rd term + d = 1st term + 3d
Observe that:
nth term = 1st term + (n-1)d  

In order to prove this by induction,
When n=1
1st term ≟ 1st term + (1-1)d
1st term = 1st term + 0
1st term = 1st term  TRUE

We let n=k
kth term = 1st term + (k-1) d , we know that this is true

But is this true for n=k+1?
(k+1)th term ≟  1st term + (k+1-1)d
(k+1)th term ≟  1st term + kd
We know for a fact that (k+1)th term = kth term + d
kth term + d 
≟ 1st term + kd
Using the equation above,
1st term + (k-1)d + d 
≟  1st term + kd
≟  kd
kd = kd  TRUE
2 5 2
Thnks sir , but how can I get the Derivation on this problem ? I didnt get it sir.
Oh, just simply check the pattern of the terms. Then we have a formula. But we are not sure if the pattern will still be true in the other terms so we prove by induction.
a_1,a_1+d,a_2+d,...,a_{n-1}+d \\ a_1+d,a_1+2d,a_1+3d,...,a_n(n-1)d
i didnt understand the sequence sir.
Oh okay so it is like this,