A well digging company charges Rs. 1,250 for the first meter, Rs. 1500 for thesecond meter and Rs. 1,750 for the third meter and so on. What is the depth of a wellthat costs Rs. 50,000

The cost of digging the well increases by Rs. 250 for each additional meter. This forms an arithmetic progression (AP) with the first term a = Rs. 1250 and the common difference d = Rs. 250.

The sum S of the first n terms of an AP is given by the formula:

S = n/2 * (2a + (n-1)d)

We know that S = Rs. 50,000. We can substitute the values of S, a, and d into the formula and solve for n:

50,000 = n/2 * (2*1250 + (n-1)*250)

Solving this equation will give us the depth of the well in meters.

Let's simplify the equation:

50,000 = n/2 * (2500 + 250n - 250)

50,000 = n/2 * (2250 + 250n)

Multiply through by 2 to get rid of the fraction:

100,000 = n * (2250 + 250n)

Expand the brackets:

100,000 = 2250n + 250n^2

Rearrange to form a quadratic equation:

250n^2 + 2250n - 100,000 = 0

Divide through by 250 to simplify:

n^2 + 9n - 400 = 0

This is a quadratic equation in the form ax^2 + bx + c = 0. We can solve it using the quadratic formula:

n = [-b ± sqrt(b^2 - 4ac)] / (2a)

n = [-9 ± sqrt((9)^2 - 41(-400))] / (2*1)

n = [-9 ± sqrt(81 + 1600)] / 2

n = [-9 ± sqrt(1681)] / 2

n = [-9 ± 41] / 2

We have two possible solutions: n = (41 - 9) / 2 = 16 or n = (-41 - 9) / 2 = -25

Since the depth of the well cannot be negative, we discard the negative solution. Therefore, the depth of the well is 16 meters.