CPhill

Let b be the second term....so we have.....

3,  b,  3 + b, 3 + 2b, 6 + 3b

And the last term is equal to 3 ...so.....

6 + 3b = 3    

3b = -3     →   b = -1

And ths series is :

3, -1, 2, 1, 3

Let n be the first integer...and we have

n + (n + 1) + (n +2) +  ..... + (n + 38) + (n +39)

40n + [(39)(40)/2]  = 100

40n + 780  = 100    subtract 780 from both sides

40n = -680    divide both sides by 40

n = -17

We have the following system

9 = a₁ + d(5 -1)

-84 = a₁ + d(32 -1)       simplifying, we have

9 = a₁ + 4d

-84 = a₁ + 31d    subtract the second equation from the first

93 = -27d     divide both sides by -27

d = -31/9

Using the first equation to find a₁, we have

9  = a₁ + 4(131/9) 

a₁ = 9 - 4(-31/9) = 205/9    and this is the first term

So....the 23rd term is given by

205/9 + (-31/9)(23-1) = -53

There are [157 - 13] / 16 + 1  = 9 + 1 = 10 terms in the sequence

The mean will be the average of the middle two terms...so we have...

[(13 + 4*16) + (13 + 5*16)] / 2  = 85

[Notice that we could have found the same result by adding the first and last terms and dividing by 2 ]

The sequence is given by

a_{n =}  a₁ -3(n - 1)    where a₁ is the first term and a_n is the nth term.....so we have......

-17 = 88 - 3(n - 1)      simplify

-17 = 88 - 3n + 3

-17  = 91 - 3n      subtract 91 from both sides

-108 = -3n        divide both sides by -3

36 = n   ....so...  35 terms appear before -17 

9 is the 1st term and 4 is the third

And the nth term is given by....    a₁(r)^(n-1 )    where a₁ is the first term....and r is the geometric multiplier.....so, we have

9(r)^(3- 1)  = 9(r)^(2)  =  4      divide both sides by  9

(r)^2 =  4/9    take the square root of both sides

r  = 2/3

So, b =  9(2/3)^(4 -1)  = 9(2/3)^3 =  8/3

Let a₁ be the first term....then the successive terms are

3a₁ + 4,  9a₁ + 16,  27a₁ + 52,  81a₁ + 160     

So .... 81a₁ + 160 = 403     subtract 160 from each side

81a₁ = 243    divide both sides by 81

a₁ = 3

The sum of the first n even integers is given by  (n)(n + 1)

The sum of the first n odd integers is given by n^2

So we have

(400)(401) - (400^2)   =

(400)(400 + 1)  - (400^2)  =

400^2 + 400 - 400^2   = 400

1/2 , 2, 5, 11, 23, 47, 95

The next term is found by doubling the previous term and adding 1  =   2(95) + 1   = 191

If A is at 13 and G is at 59......46 equal intervals separate them....and these are broken into 6 equal parts. Thus each of these parts span 46/6 = 23/3 integers.

And the sum of the numbers at B,C, D, E and F is given by

13(5) + (23/3)[(5)(6)/2]=  180