4(N-3) is the formula, do the math from there and you get "N-3"
3 (n + 7) = 16
7(n+3)
6(3n+5)
3*(n + 5) or 3n + 15
sum = 2n + 3m
4(N-3) is the formula, do the math from there and you get "N-3"
10n = -30 n = -3
To find the (mean) average, add all the numbers and divide by the number of numbers. The sum of a series of digits (in arithmetic progression, like 13, 14, 15, ... 37) is sum = (first + last) x number_of_digits / 2 So their average is: average = ((first + last) x number_of_digits / 2) / number_of_digits = (first + last) / 2 = average of first and last digits! So the average of the numbers 13, 14, 15, ..., 37 is: average = (13 + 37) / 2 = 50 / 2 = 25 To find the sum of n digits starting with m: Sum = m + (m+1) + ... + (m+n-2) + (m+n-1) Rewrite the sum in reverse order: Sum2 = Sum = (m+n-1) + (m+n-2) + ... + (m+1) + m Add the two sums, term by term: Sum + Sum2 = 2 Sum = (m + (m+n-1)) + (m + (m+n-1)) + ... + (m + (m+n-1)) + (m + (m+n-1)) There are n terms, all (m + (m+n-1)), so: 2 Sum = (m + m+n-1) x n Sum = (m + m+n-1) x n / 2 But n is the number of digits, m is the first number and (m+n-1) is the last, so: Sum = (first + last) x number_of_digits / 2
m and n are 70 and 90
3 (n + 7) = 16
7(n+3)
Where n = any number, n(12+3)
4*(n + 1) = 6*n*3
m -5 n-3
The sum of the 1st n terms is : N(3N-1)/2 Explanation : The sum from 1 to N of (3m-2) = 3 * sumFrom1toN(m) - sumFrom1toN(2) = 3 * (N*(N+1)/2) -2*N = N(3N-1)/2 For N=10 => 145
3(n+2) = -8 n = -14/3 or 4 and 2/3