Need urgent maths help..!!

http://postimage.org/image/1tbf9hjxg/
I need answers of 7th and 8th Qs (section b) of the link provided.
If you find the answer too long you may contact me through my profile. 10 pts guaranteed.

PS : this is not my homework

A.P. with a = 20, d = -2/3
let the # of terms be n, then
last term = a + (n-1)d = 20 +(n-1)*(-2/3)
and sum of terms = (n/2)(2a + (n-1)d)
20n + (n/2)(n-1)(-2/3) = 300
60n - n(n-1) = 900
n^ - 61n + 900 =0
(n - 25)(n-36) = 0
n = 25 or 36
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the equation is parabolic,
and the sum of the terms from n = 26 to 36 (inclusive) is zero
so we get 2 values for n

The common difference is -2/3.

So, we have Sn=300, n=?, a=20 and d=-2/3

Then, by the formula Sn=n/2 [2a+(n-1)d]

We have
300=n/2 (40+[n-1]*-2/3)
600=n (40-2/3n+2/3)
600=n ([120+2]/3 -2n/3)
600= 2n/3 (60+1-n)
900= n (61-n)
900= 61n - n^2

Changing sides, we get

n^2 - 61n+900 = 0

n^2 -25n-36n+900=0
n (n-25) -36 (n-25)=0
(n-36)(n-25)=0

n=36 or 25

The double answer arises because of the quadratic equation, which has two solutions. The value of terms are on either side of zero, thus cancelling out some of them.

Hope this helped.