Help please

[Joyce]

This question isn't that hard but I can't seem to get the same anser as the back of the book:

Determine the sum of this arithmetic series:

20+14+8+.......-40


This is what I did:

S= 20+14+8+......-40
S=-40-34-28-........+20

6S= -20-20-20.........-20

6S=60x(-20)

=-200

 

[stapel]

I must not be understanding what you're doing, because it looks to me like you're saying that S + S equals 6S...?

Please reply with clarification. Thank you.

Eliz.

 

[TchrWill]

Determine the sum of this arithmetic series:

20+14+8+.......-40


This is what I did:

S= 20+14+8+......-40
S=-40-34-28-........+20

6S= -20-20-20.........-20

6S=60x(-20)

It appears you attempted to take the route that Gauss took. Somehow you went astray with 6S = 60x(-20)

Starting over:

S= 20+14+8+......-40 (11 terms)
S=-40-34-28-......+20 (11 terms)

Adding:

2S = 11(-20) = -220 making S = -110.

You could also use the standard formula for the sum of an arithmetic progression.
The sum of any arithmetic progression derives from S = n(a + L)/2 where a = the first term, L = the last term and n = the number of times.

Therefore, S = 11(20 - 40)/2 = -110.

 

[Joyce]

Thank you very much for your help, I see where I went wrong now.

 

[Denis]

20 + 14 + 8 + ....... -40

The number of terms n, 11 in this case, can be obtained this way (d = difference):
n = (L - a) / d + 1
n = (-40 - 20) / -6 + 1
n = -60 / -6 + 1
n = 10 + 1 = 11

Or rewrite the series:
-40 + -34 + -28 + ..... 20
n = (20 - (-40)) / 6 + 1
n = (20 + 40) / 6 + 1
n = 60 / 6 + 1
n = 10 + 1 = 11

So we can substitute n = (L - a) / d + 1 in the Sum equation S = n(a + L) / 2:
S = [(L - a) / d + 1][(a + L)/2]
simplifies to:
S = (a + L)(L - a + d) / (2d)
In other words, no need to worry about number of terms...

Using -40 + -34 + -28 + ..... 20 :
S = (a + L)(L - a + d) / (2d)
S = (-40 + 20)(20 - (-40) + 6) / (2*6)
S = (-20)(66) / 12
S = -1320 / 12
S = -110