Re: series, because our stupid teacher is stupid :). As is margaret groves. (1 Viewer)

[stuka]

Lol, any1 know how to do the partial sum of a series if the common difference is negative. E.g

How many terms of the series 80+73+66... give a sum of 495.

I tried:

495=n/2(2a+(n-1))
nd it didnt work, is my algebra really crap or is there a way to do it?
Thankyou!!!!

 

[Drongoski]

Lol, any1 know how to do the partial sum of a series if the common difference is negative. E.g

How many terms of the series 80+73+66... give a sum of 495.

I tried:

495=n/2(2a+(n-1)d)
nd it didnt work, is my algebra really crap or is there a way to do it?
Thankyou!!!!

.: 495 = n/2(2x80 +(n-1)x(-7)) ==> n(167 -7n) = 990

.: 7n² - 167n + 990 = 0

i.e. (7n-90)(n-11) = 0 ==> n = 11

 

[stuka]

haha cheers, yeah i used -7 forgot to put that in, i guess my algebra is just rubbish . Thankyou!!!

 

[4025808]

Lol, any1 know how to do the partial sum of a series if the common difference is negative. E.g

How many terms of the series 80+73+66... give a sum of 495.

I tried:

495=n/2(2a+(n-1))
nd it didnt work, is my algebra really crap or is there a way to do it?
Thankyou!!!!

since 'a' is the indication of the first term, and that 80 is the first term,
let a = 80
and that -7 is the common difference

now use the AP sum formula

n/2(2x80 +(n-1)-7) = 495
n(160 - 7n + 7) = 990
160n -7n^2 + 7n = 990
7n^2 -167n +990 = 0
(7n-90)(n-11)=0

n = 11 or 90/7
but because it can only be an integer, therefore n = 11