Find the sum of all positive 3-digit numbers divisible by 12.

First 3-digit number divisible by 12 is 108.

Last 3-digit number divisible by 12 is 996.

108 = 12 · 9

996 = 12 · 83

From 108 to 996 are 83 - 9 + 1 = 75 numbers divisible by 12.

108 is the first term of the arithmetic sequence.

996 is the 75th term of the arithmetic sequence.

Use the formula of a sum of terms of an arithmetic sequence:

We have:

Substitute:

cant

step-by-step explanation: there is no picture

x coordinate can be written by using the formula, x= x1 + 2/5(x2 - x1)

step-by-step explanation:

formula

x = x1 + p/w(x2 - x1)

let the line segment ab with a(x1,y1) and b(x2,y2)

a point(x,y) divide ab in the ratio, 2: 3 the x coordinate can be written as

x = x1 + p/w(x2 - x1)

here p=2 and w = 2+3

therefore,

x coordinate can be written by using the formula, x= x1 + 2/5(x2 - x1)

similarly y coordinate y = y1 + 2/5(y2 - y1)