Halp!!!!!

A certain 300-term geometric sequence has first term 1337 and common ratio of -1/2. How many terms of this sequence are greater than 1?

Note that the terms that are positive are the odd terms ... 

And each positive term  can  actually be written as

1337 (1/4)^n   where n is the nth positive term  after the first one

And we want to find this

1337 (1/4)^n > 1

(1/4)^n  > 1/1337       take the log of each side

log (1/4)^n > log (1/1337)

And we can write......[remember to reverse the sign since  log (1/4)  is negative ]

n < log (1/1337)/log(1/4)  

n < 5.19

So....there are 5 terms [after the first one] > 1   .....which gives 6 terms total

Proof ....the 10th term after the first one - the 6th positive one-  is     1337(-1/2)^(10) ≈ 1.03

And the 12th term after the first one - the 7th positive one -  is  1337(-1/2)^(12) ≈ .326

Laverne starts counting out loud by 5's. She starts with 7. As Laverne counts, Shirley sums the numbers Laverne says. When the sum finally exceeds 5000, Shirley runs screaming from the room. What number did Laverne last say before Shirley flees?

Note that we have this series

                                                                 5n + 7 - 7 +1  

7, 12, 17, 22, 27, 32.....

Note that the first term is 7  and the last is  5(n-1) + 7 =  5n + 2

And the sum  can  be represented as  [ first term + last term ] * number of terms / 2

So we  need to solve this

[ 7 + 5n + 2] * n / 2  > 5000        simplify this

[5n + 9 ] * n > 10000

5n^2 + 9n > 10000

5n^2 + 9n - 10000 > 0

The positive solution for this is when  n > 43.83

Note that the 43rd term 5(43) + 2  = 217 

And the sum of the first 43 terms is  [ 7 + 217] * 43/2  = 4816 

And  the  44th term is 222

And this will cause the sum to be greater than 5000

So....before  Shirley flees, the last number Laverne says is 217

Let a_1, a_2, . . . , a_{10} be an arithmetic sequence. If a_1 + a_3 + a_5 + a_7 + a_9 = 17 and a_2 + a_4 + a_6 + a_8 + a_{10} = 15, then find a_1.

Letting d be the common difference between terms, note that we can write the terms of the first sum as

[a1] + [a1 + 2d] + [a1 + 4d ] + [ a1 + 6d ] + [a1 + 8d ]  = 17

Simplify as .... 5a1 + 20d  = 17    (1)

And for the sum of the second series, we can write the terms as

[a1 + d ] + [a1 + 3d ] + [ a1 + 5d ] + [ a1 + 7d] + [ a1 + 9d]  = 15

Simplify as ....  5a1 + 25d  = 15   (2) 

Subtract  (1)  from (2)  and we have that

5d  = -2

d  =  -2/5

And using (1) to solve for  a1

5a1 + 20 (-2/5) = 17

5a1  - 8  =  17

5a1  = 25

a1  = 25/5 = 5

Check this with (2)

5(5) + 25(-2/5)  =

25 - 10  =

15