The sum of the first ten terms of an arithmetic progression consisting of positive integer terms is equal to the sum of the 20th, 21st, 22nd term. If the first term is less than 20, find how many terms are required to give a sum of 960?

30 units^2

step-by-step explanation:

area is l × w

the length is 3

and the width is 10

3 × 10= 30 units squared

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.

step-by-step explanation:

the distance (d) between two points (x1,y1) and (x2,y2) is given by the formula

d = √   ((x2-x1)2+(y2-y1)2)

d = √ (-11-6)2+(4-3)2

d = √ )2+(1)2)

d = √ (289+1)

d = √ 290

the distance between the points is 17.0293863659264

the midpoint of two points is given by the formula

midpoint= ((x1+x2)/2,(y1+y2)/2)

find the x value of the midpoint

xm=(x1+x2)/2

xm=(6+-11)/2=-2.5

find the y value of the midpoint

ym=(y1+y2)/2

ym=(3+4)/2=3.5

the midpoint is: (-2.5,3.5)

graphing the two points, midpoint and distance

p1 (6,3)

p2 (-11,4)

midpoint (-2.5,3.5)

the length of the black line is the distance between the points (17.0293863659264)

find the slope of the line connecting the two points

slope =   (y2-y1)   =   (4-3)   =   (1)   = -0.0588235294117647

(x2-x1) (-11-6) (-17)

find the equation of the line passing through the two points

the general equation for a straight line is

y = mx + b

where m represents the slope of the line which we found in the previous step to be -0.0588235294117647

y = -0.06x + b

we substitute x and y for the values from one of our points (6,3)

3 = -.06×6 + b

3 = -0.35 + b

3--0.35 = b

3.35 = b

knowing both b and m, we can contruct the equation of the line

y= -0.06x+ 3.35

x and y intercepts

the x-intercept is a point on the graph where y is zero

using the equation we found in the previous step and substituting zero for y

y= -0.06x+ 3.35

0= -0.06x+ 3.35

0.06x= 3.35

x= 3.35/0.06 = 57.00

the x intercept for this straight line is 57.00

the y-intercept is a point on the graph where x is zero

using the equation we found in the previous step and substituting zero for x

y= -0.06×0+ 3.35

y= 3.35

the y intercept for this straight line is 57.00

the distance (d) between two points (x1,y1) and (x2,y2) is given by the formula

d = √   ((x2-x1)2+(y2-y1)2)

d = √ (-11-6)2+(4-3)2

d = √ )2+(1)2)

d = √ (289+1)

d = √ 290

the distance between the points is 17.0293863659264

the midpoint of two points is given by the formula

midpoint= ((x1+x2)/2,(y1+y2)/2)

find the x value of the midpoint

xm=(x1+x2)/2

xm=(6+-11)/2=-2.5

find the y value of the midpoint

ym=(y1+y2)/2

ym=(3+4)/2=3.5

the midpoint is: (-2.5,3.5)

graphing the two points, midpoint and distance

p1 (6,3)

p2 (-11,4)

midpoint (-2.5,3.5)

the length of the black line is the distance between the points (17.0293863659264)

find the slope of the line connecting the two points

slope =   (y2-y1)   =   (4-3)   =   (1)   = -0.0588235294117647

(x2-x1) (-11-6) (-17)

find the equation of the line passing through the two points

the general equation for a straight line is

y = mx + b

where m represents the slope of the line which we found in the previous step to be -0.0588235294117647

y = -0.06x + b

we substitute x and y for the values from one of our points (6,3)

3 = -.06×6 + b

3 = -0.35 + b

3--0.35 = b

3.35 = b

knowing both b and m, we can contruct the equation of the line

y= -0.06x+ 3.35

x and y intercepts

the x-intercept is a point on the graph where y is zero

using the equation we found in the previous step and substituting zero for y

y= -0.06x+ 3.35

0= -0.06x+ 3.35

0.06x= 3.35

x= 3.35/0.06 = 57.00

the x intercept for this straight line is 57.00

the y-intercept is a point on the graph where x is zero

using the equation we found in the previous step and substituting zero for x

y= -0.06×0+ 3.35

y= 3.35

the y intercept for this straight line is 57.00 and therefore the awnser is c.