Here are the first five terms of an arithmetic sequence. 7
13
19
25
31
Prove that the difference between the squares of any two terms of the sequence is always
a multiple of 24.

9514 1404 393

  the difference of squares is 24(3n+2), so a multiple of 24

Step-by-step explanation:

The sequence has a first term of 7 and a common difference of 6, so the n-th term is ...

  a[n] = a1 +d(n -1)

  a[n] = 7 + 6(n -1)

  a[n] = 6n +1

Then the (n+1)-th term is ...

  a[n+1] = 6(n+1) +1 = 6n +7

The difference of the squares of these terms is ...

  a[n+1]^2 -a[n]^2 = (6n+7)^2 -(6n+1)^2

  = (36n^2 +84n +49) -(36n^2 +12n +1)

  = 72n +48

  = 24(3n +2) . . . . . always a multiple of 24

The difference of squares of terms n and n+1 is 24(3n+2), a multiple of 24.

Check

The difference between squares for n=1 and n=2 is 13^2 -7^2 = 169 -49 = 120. The formula tells us the difference is 24(3·1 +2) = 24·5 = 120, in agreement.

29 mil

step-by-step explanation:

see attached

step-by-step explanation:

given the function as;

write the equation replacing f(x) with y

form   and fill a table containing values of y with corresponding values of x

you can take a range for   values of x as -3≤x≤3

solving for x=-3

for x=-2

repeat the same procedure to get the values of y for the remaining ,x=-1 to x=3

plot the values of y against values   of x and obtain the following graph