Stress = Young's modulus × Strain = 150 GPa × 0.01 = 150 MPa.

According to Hooke's Law, if stress is doubled, strain also doubles for elastic materials.

The sum S of an infinite GP is given by S = a / (1 - r). Here, 20 = a / (1 - 1/4) = a / (3/4). Thus, a = 20 * (3/4) = 15.

The sum of an infinite GP is given by S = a / (1 - r). Here, S = 10 and r = 1/3. Thus, 10 = a / (1 - 1/3) = a / (2/3) => a = 10 * (2/3) = 20.

Let the son's age be x. Then father's age = 4x. According to the problem, x + 4x = 50, which gives 5x = 50, so x = 10.

A triangle with one angle measuring 90 degrees is classified as a right triangle, as it adheres to the property of having one angle equal to 90 degrees.

In an equilateral triangle, each angle is equal, so each angle = 180/3 = 60 degrees.

The third angle can be found by subtracting the sum of the other two angles from 180 degrees: 180 - (50 + 70) = 60 degrees.

A triangle with one angle measuring 90 degrees is defined as a right triangle.

Let the two-digit number be 10a + b, where a is the tens digit and b is the units digit. We have a + b = 9 and 10a + b = 4(a + b). Solving these gives a = 4 and b = 5, so the number is 45.

The factors of 12 are 1, 2, 3, 4, 6, and 12. Their sum is 28.

Using the formula for the sum of a geometric series, S_n = a(1 - r^n) / (1 - r), we can solve for r. Here, S_5 = 1(1 - r^5) / (1 - r) = 31, leading to r = 3.

The sum S_5 = 5/2 * (2a + 4d). Here, 50 = 5/2 * (2a + 8), solving gives a = 10.

The average of the first n terms is given by S_n/n. Here, S_5 = 50, so the average = 50/5 = 10.

The sum of the first n natural numbers is given by the formula n(n + 1)/2 = 210. Solving n(n + 1) = 420, we find n = 14 or n = 15. Thus, n = 15.

The sum of the first n natural numbers is given by the formula n(n + 1)/2 = 255. Solving n(n + 1) = 510, we find n = 16.

The sum of the first n natural numbers is given by the formula n(n + 1)/2. Setting this equal to 2550, we have n(n + 1) = 5100. Solving the quadratic equation n^2 + n - 5100 = 0 gives n = 50.

The sum of the first n natural numbers is given by the formula n(n + 1)/2. Setting this equal to 5050, we have n(n + 1)/2 = 5050. Solving for n gives n = 100.

n(n + 1)/2 = 55, solving gives n = 10.

Using the formula S_n = a(1 - r^n) / (1 - r), we have 63 = 3(1 - r^4) / (1 - r). Solving gives r = 2.

The sum S_n can be negative if the common ratio r is negative and the first term a is also negative.

As n approaches infinity and |r| < 1, r^n approaches 0, thus S_n approaches a/(1-r).

Using the formula S_n = a(1 - r^n) / (1 - r), we have 81 = 3(1 - r^n) / (1 - r). Solving gives r = 3.

When r = 1, S_n = a(1 - 1^n)/(1 - 1) is indeterminate, but the sum of n terms is na.

Using the formula S_n = a(1 - r^n) / (1 - r), we have 63 = 3(1 - r^4) / (1 - r). Solving gives r = 2.

Using the formula for the sum of a GP, S_n = a(1 - r^n) / (1 - r), we have 63 = 7(1 - 2^n) / (1 - 2). Solving gives n = 5.

The sum of the first n terms S_n = n/2 * (2a + (n-1)d). By differentiating S_n with respect to n, we can find the common difference. The common difference is 3.

The common difference can be found by calculating S_n - S_(n-1). S_n = 3n^2 + 2n and S_(n-1) = 3(n-1)^2 + 2(n-1). Simplifying gives the common difference as 6.

The 4th term a_4 = S_4 - S_3 = (3(4^2) + 2(4)) - (3(3^2) + 2(3)) = (48 + 8) - (27 + 6) = 56 - 33 = 23.

The common difference can be found by calculating S_n - S_(n-1). Here, S_n = 3n^2 + 2n and S_(n-1) = 3(n-1)^2 + 2(n-1). The difference simplifies to 4.