Given: Sum (Sn) of n-terms of an Arithmetic Progression is (2n^2 + n).

To find: Difference of its 10th and 1st terms.

Solution:

1. Formula for Sum of n terms of AP:
The sum of n terms of an Arithmetic Progression is given by:
Sn = n/2(2a + (n-1)d), where a is the first term, d is the common difference and n is the number of terms.

2. Given Sum of n terms of AP:
Here, the Sum of n terms of an Arithmetic Progression is (2n^2 + n).
Therefore, Sn = 2n^2 + n.

3. Using the formula of Sn:
We know that the sum of n terms of AP is given by:
Sn = n/2(2a + (n-1)d)

Substituting the given value of Sn in the formula, we get:
2n^2 + n = n/2(2a + (n-1)d)

Simplifying the equation, we get:
4a + 2nd = 2n + 1

4. Finding the first term:
We know that a = (2Sn - n^2)/2n
Substituting the given value of Sn in the formula, we get:
a = (2(2n^2 + n) - n^2)/2n
a = (3n^2 + n)/2

Therefore, the first term a = (3n^2 + n)/2.

5. Finding the common difference:
From equation (4), we have:
4a + 2nd = 2n + 1

Substituting the value of a from equation (4), we get:
4[(3n^2 + n)/2] + 2nd = 2n + 1

Simplifying the equation, we get:
3d = 4

Therefore, the common difference d = 4/3.

6. Finding the 10th term:
We know that the nth term of an AP is given by:
an = a + (n-1)d

Substituting the values of a and d, we get:
a10 = (3n^2 + n)/2 + (10-1)(4/3)
a10 = (3n^2 + n)/2 + 36/3
a10 = (3n^2 + n + 36)/2

Substituting n = 10, we get:
a10 = (3(10^2) + 10 + 36)/2
a10 = 196/2
a10 = 98

Therefore, the 10th term of the AP is 98.

7. Finding the 1st term:
We know that a = (3n^2 + n)/2.

Substituting n = 1, we get:
a1 = (3(1^2) + 1)/2
a1 = 2

Therefore, the 1st term of the AP is 2.

8. Finding the difference between 10th and 1st terms:
The difference between the 10th and 1st terms is given by:
d10-1 = a10 - a1
d10-1 = 98 -