An arithmetic progression (AP) is a sequence of numbers in which each term is obtained by adding a constant difference (d) to the preceding term. In other words, an AP is a sequence of numbers in which each term differs from the preceding term by a fixed number called the common difference. The formula for the nth term of an AP is given by: an = a1 + (n-1)d Where, an = nth term of the AP a1 = first term of the AP n = number of terms in the AP d = common difference Examples of arithmetic progressions: In this AP, the first term (a1) is 2 and the common difference (d) is 2. The second term is obtained by adding 2 to the first term (a2 = a1 + d = 2 + 2 = 4). Similarly, the third term is obtained by adding 2 to the second term (a3 = a2 + d = 4 + 2 = 6), and so on. In this AP, the first term (a1) is -3 and the common difference (d) is 4. The second term is obtained by adding 4 to the first term (a2 = a1 + d = -3 + 4 = 1). Similarly, the third term is obtained by adding 4 to the second term (a3 = a2 + d = 1 + 4 = 5), and so on. In this AP, the first term (a1) is 10 and the common difference (d) is -3. The second term is obtained by adding -3 to the first term (a2 = a1 + d = 10 – 3 = 7). Similarly, the third term is obtained by adding -3 to the second term (a3 = a2 + d = 7 – 3 = 4), and so on. Sum of N Terms in AP with Examples The sum of the first n terms of an arithmetic progression (AP) is called the sum of n terms or the nth partial sum of the AP. The formula for finding the sum of the first n terms of an AP is given by: Sn = (n/2)(a1 + an) Where, Sn = sum of first n terms of the AP a1 = first term of the AP an = nth term of the AP We can also write the formula for the nth term of an AP in terms of the common difference (d): an = a1 + (n-1)d Using this formula, we can rewrite the formula for the sum of n terms as: Sn = (n/2)(2a1 + (n-1)d) Examples: Here, a1 = 2 and d = 2 Using the formula, Sn = (n/2)(2a1 + (n-1)d) = (10/2)(2(2) + (10-1)(2)) = 110 So, the sum of the first 10 terms of the AP is 110. Here, a1 = -3 and d = 4 Using the formula, Sn = (n/2)(2a1 + (n-1)d) = (15/2)(2(-3) + (15-1)(4)) = 120 So, the sum of the first 15 terms of the AP is 120. Note that the formula for the sum of n terms of an arithmetic progression can be used only when the number of terms is known. If the number of terms is not known, we can use other methods to find the sum, such as using the sum of the first and last terms, or using the sum of two APs with the same first term and common difference.

Vedic Maths is an ancient Indian system of mathematics that is based on 16 sutras, or word-formulas, and 13 sub-sutras, or corollaries. One of the most well-known Vedic Maths tricks is for finding the square root of a number. Here is how it works: Step 1: Find the largest square that is less than or equal to the given number. Let’s call this square “A”. Step 2: Subtract A from the given number to get “B”. Step 3: Divide B by twice A to get “C”. Step 4: Add A and C to get the approximate square root of the given number. Step 5: If the approximate square root is not accurate enough, repeat steps 2-4 with the new number instead of the original number until you get the desired level of accuracy. Let’s take an example to illustrate this process. Suppose we want to find the square root of 73. Step 1: The largest square less than or equal to 73 is 64 (which is 8 squared). Step 2: 73 – 64 = 9 Step 3: 9 / (2 x 8) = 0.5625 Step 4: 8 + 0.5625 = 8.5625 (approximate square root) Step 5: If we want more accuracy, we can repeat steps 2-4 with the new number 9 instead of 73. This Vedic Maths trick can be quite useful for finding square roots quickly, especially for large numbers that are not perfect squares.