The sum of the sixth and eighth terms of an arithmetic progression is 40. The seventh term of this progression is 4times the second term a. Find the first term and the common difference b. The least number of terms required for the sum of thr progression to exceed 1000

"The sum of the sixth and eighth terms of an arithmetic progression is 40."
The sixth term is a₆ = a + (6-1)d = a + 5d.
The eighth term is a₈ = a + (8-1)d = a + 7d.
So, a₆ + a₈ = 40
(a + 5d) + (a + 7d) = 40
2a + 12d = 40
Divide by 2:
a + 6d = 20 (Equation 1)

"The seventh term of this progression is 4 times the second term."
The seventh term is a₇ = a + (7-1)d = a + 6d.
The second term is a₂ = a + (2-1)d = a + d.
So, a₇ = 4 × a₂
a + 6d = 4(a + d)
a + 6d = 4a + 4d
Rearrange the terms to group 'a' and 'd':
6d - 4d = 4a - a
2d = 3a (Equation 2)

From Equation 2, we can express a in terms of d:
a = (2/3)d

Substitute this expression for a into Equation 1:
(2/3)d + 6d = 20
To add the 'd' terms, find a common denominator:
(2/3)d + (18/3)d = 20
(20/3)d = 20
Multiply both sides by (3/20):
d = 20 × (3/20)
d = 3

Now substitute the value of d back into Equation 2 to find a:
3a = 2(3)
3a = 6
a = 6 / 3
a = 2

The first term is 2 and the common difference is 3.

The sum of the first n terms is Sₙ = (n/2)[2a + (n-1)d].
We need Sₙ > 1000.
Substitute a = 2 and d = 3 into the formula:
(n/2)[2(2) + (n-1)3] > 1000
(n/2)[4 + 3n - 3] > 1000
(n/2)[3n + 1] > 1000

Multiply both sides by 2:
n(3n + 1) > 2000
3n² + n > 2000
3n² + n - 2000 > 0

To find the values of n that satisfy this inequality, we first find the roots of the quadratic equation 3n² + n - 2000 = 0 using the quadratic formula:
n = [-b ± √(b² - 4ac)] / (2a)
Here, a=3, b=1, c=-2000.
n = [-1 ± √(1² - 4 × 3 × -2000)] / (2 × 3)
n = [-1 ± √(1 + 24000)] / 6
n = [-1 ± √(24001)] / 6

Calculate the square root:
√(24001) ≈ 154.9225

Now find the two possible values for n:
n₁ = (-1 + 154.9225) / 6 = 153.9225 / 6 ≈ 25.65
n₂ = (-1 - 154.9225) / 6 = -155.9225 / 6 ≈ -25.99

Since n represents the number of terms, it must be a positive integer. The inequality 3n² + n - 2000 > 0 holds for n values greater than the positive root.
So, n > 25.65.

Since n must be an integer and greater than 25.65, the smallest integer value for n is 26.

Let's check:
For n = 25: S₂₅ = (25/2)[2(2) + (25-1)3] = (25/2)[4 + 24 × 3] = (25/2)[4 + 72] = (25/2)[76] = 25 × 38 = 950. (This is not > 1000)
For n = 26: S₂₆ = (26/2)[2(2) + (26-1)3] = 13[4 + 25 × 3] = 13[4 + 75] = 13[79] = 1027. (This is > 1000)

The least number of terms required for the sum of the progression to exceed 1000 is 26.