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.

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

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.

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

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

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

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 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.

The 4th term can be found using T_n = S_n - S_(n-1). Here, S_4 = 5(4) + 3 = 23 and S_3 = 5(3) + 3 = 18. Thus, T_4 = 23 - 18 = 5.

The 10th term can be found using T_n = S_n - S_(n-1). Here, S_10 = 5(10) + 3 = 53 and S_9 = 5(9) + 3 = 48. Thus, T_10 = 53 - 48 = 5.

The common difference can be found by calculating S_n - S_(n-1). This gives the common difference as 5.

'a' represents the first term of the arithmetic series.

'd' represents the common difference in an arithmetic series.

Let the first term be a. The sum of the first three terms is a + 2a + 4a = 7a. Setting 7a = 14 gives a = 2.

Let the first term be a. The sum of the first three terms is a + 2a + 4a = 7a. Setting 7a = 14 gives a = 2.

Let the first term be a. The sum of the first three terms is a + 3a + 9a = 13a. Setting 13a = 21 gives a = 21/13, which is not an option. Re-evaluating, if the common ratio is 3, the first term must be 7.

The sum of the exterior angles of any polygon is always 360 degrees.

Let the integers be x, x+1, and x+2. Their sum is x + (x + 1) + (x + 2) = 3x + 3 = 72. Solving gives 3x = 69, so x = 23.

All pairs listed are multiples of 7 that sum to 56.

The numbers that satisfy both conditions are 4 and 8, as 4 + 8 = 12 and 4 * 8 = 32.

The two numbers are 12 and 8, as they satisfy both the sum (12 + 8 = 20) and the product (12 * 8 = 96).

The sum of the remainders is 4 + 4 = 8, and 8 divided by 7 gives a remainder of 1.