Find the sum of the 20 terms in an arithmetic progression whose first term is 7 and last term 117?

An arithmetic progression is a sequence of numbers in which the difference between any two consecutive terms is constant. Let's call this constant difference "d". We know that the first term of the sequence is 7 and the last term is 117. Let's call the number of terms in the sequence "n". Using this information, we can write two equations: 1. The nth term of an arithmetic progression formula: an = a1 + (n - 1)d, where "an" is the nth term of the sequence, "a1" is the first term of the sequence, "n" is the number of terms in the sequence, and "d" is the constant difference between any two consecutive terms in the sequence. We can use this equation to find the value of "d": 117 = 7 + (n - 1)d 110 = (n - 1)d d = 110/(n-1) 2. The sum of an arithmetic progression formula: Sn = n/2 * (a1 + an), where "Sn" is the sum of the first "n" terms of the sequence. We can use this equation to find the sum of the first 20 terms of the sequence: Sn = 20/2 * (7 + (7 + 19d)) Sn = 10 * (14 + 2090/(n-1)) Now we can substitute the value of "d" from equation (1) into equation (2) and simplify: Sn = 10 * (14 + 2090/(n-1)) Sn = 10 * (14 + 2090/(20-1)) (since there are 20 terms in the sequence) Sn = 10 * (14 + 110) Sn = 1240 Therefore, the sum of the 20 terms in the arithmetic progression is 1240