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Please notice that an arithmetic sequence looks like this:

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

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

(k-1+1)d ≟ kd

kd = kd

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

(k-1+1)d ≟ kd

kd = kd

*TRUE*