**Explanation:**

To find the expression for Sn, let's start by writing out the first few terms of the arithmetic progression:

a, a + d, a + 2d, a + 3d, ...

From this, we can observe that the nth term of the arithmetic progression can be expressed as:

an = a + (n-1)d

Now, let's find the sum of the first n terms, Sn:

Sn = a + (a + d) + (a + 2d) + ... + [a + (n-1)d]

To simplify this expression, we can rearrange the terms:

Sn = (a + a + a + ... + a) + (d + d + d + ... + d) + (2d + 2d + ... + 2d) + ... + [(n-1)d + (n-1)d + ... + (n-1)d]

Notice that we have n terms in each group, and the sum of n identical terms is n times the value of that term. Therefore, we can rewrite Sn as:

Sn = (n * a) + (n * d) + (n * 2d) + ... + (n * (n-1)d)

Factoring out n, we get:

Sn = n * [a + (a + d) + (a + 2d) + ... + (a + (n-1)d)]

Using the formula for the sum of an arithmetic series, we can simplify this further:

Sn = n * [(n/2) * (2a + (n-1)d)]

Now, let's find Sn-1:

Sn-1 = (n-1) * [a + (a + d) + (a + 2d) + ... + (a + (n-2)d)]

Using the formula for the sum of an arithmetic series, we can simplify this expression as well:

Sn-1 = (n-1) * [(n-1)/2 * (2a + (n-2)d)]

Finally, let's find Sn-2:

Sn-2 = (n-2) * [a + (a + d) + (a + 2d) + ... + (a + (n-3)d)]

Using the formula for the sum of an arithmetic series, we can simplify this expression too:

Sn-2 = (n-2) * [(n-2)/2 * (2a + (n-3)d)]

**Answer:**

To find Sn-2Sn-1, we subtract Sn-1 from Sn:

Sn - Sn-1 = [n * (2a + (n-1)d)] - [(n-1) * (2a + (n-2)d)]

Simplifying this expression, we get:

Sn - Sn-1 = n * (2a + (n-1)d) - (n-1) * (2a + (n-2)d)

Expanding the terms, we have:

Sn - Sn-1 = 2an + nd - 2an + 2ad - (2an - 2ad + nd - 2d)

Combining like terms, we get:

Sn - Sn-1 = 2ad - 2d

Therefore, Sn - Sn-1 = 2d(a -

Sn = (n/2)[ 2a + ( n -1) d]
then Sn - 2Sn-1 + Sn +  2
⇒ (n/2)[ 2a + ( n -1) d] - 2(n - 1)/2)[ 2a + ( n - 1 -1) d] + (n +2) / 2)[ 2a + ( n + 2 -1) d]
⇒ (1/2)[ 2an + n( n -1) d] + [ 4a(n - 1) + 2(n - 1)( n - 2) d] +[ 2a(n + 2)+ ( n + 1) (n + 2) d]
⇒ (1/2)[ 2a[ n - 2n + 2 + n + 2] + d [ n2 - n - 2n2 + 6n - 4 + n2 + 3n + 2] ]
⇒ (1/2)[ 2a(4) + d(8n - 2) ]
= [ 4a + (4n - 1)d]

That's all🙂