Question 1011827: write 3 different series each with 5 terms and each with a sum of 100.
Found 3 solutions by Edwin McCravy, AnlytcPhil, Theo:
Answer by Edwin McCravy(20056)   (Show Source):
Answer by AnlytcPhil(1806)  (Show Source):
You can put this solution on YOUR website!
-10+5+20+35+50 = 100, is an arithmetic series with 1st term -10
and common difference 15.
-8+6+20+34+48 = 100, is an arithmetic series with 1st term -8
and common difference 14.
-6+7+20+33+46 = 100, is an arithmetic series with 1st term -6
and common difference 13.
-4+8+20+32+44 = 100, is an arithmetic series with 1st term -4
and common difference 12.
-2+9+20+31+42 = 100, is an arithmetic series with 1st term -2
and common difference 11.
0+10+20+30+40 = 100, is an arithmetic series with 1st term 0
and common difference 10.
2+11+20+29+38 = 100, is an arithmetic series with 1st term 2
and common difference 9.
4+12+20+28+36 = 100, is an arithmetic series with 1st term 4
and common difference 8.
6+13+20+27+34 = 100, is an arithmetic series with 1st term 6
and common difference 7.
8+14+20+26+32 = 100, is an arithmetic series with 1st term 8
and common difference 6.
10+15+20+25+30 = 100, is an arithmetic series with 1st term 10
and common difference 5.
12+16+20+24+28 = 100, is an arithmetic series with 1st term 12
and common difference 4.
14+17+20+23+26 = 100, is an arithmetic series with 1st term 14
and common difference 3.
16+18+20+22+24 = 100, is an arithmetic series with 1st term 16
and common difference 2.
18+19+20+21+22 = 100, is an arithmetic series with 1st term 18
and common difference 1.
20+20+20+20+20 = 100, is an arithmetic series with 1st term 20
and common difference 0.
22+21+20+19+18 = 100, is an arithmetic series with 1st term 22
and common difference -1.
24+22+20+18+16 = 100, is an arithmetic series with 1st term 24
and common difference -2.
26+23+20+17+14 = 100, is an arithmetic series with 1st term 26
and common difference -3.
28+24+20+16+12 = 100, is an arithmetic series with 1st term 28
and common difference -4.
30+25+20+15+10 = 100, is an arithmetic series with 1st term 30
and common difference -5.
Edwin
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! i'll use an arithmetic series because there are formulas that should allow me to figure out what numbers could be used.
Sn = 1/2 * n * (A1 + An)
n = 5, therefore:
S5 = 1/2 * 5 * (A1 + A5)
multiply both sides by 2 to get:
2 * S5 = 5 * (A1 + A5)
divide everything by 5 to get:
2 * S5 / 5 = A1 + A5
subtract A1 from both sides of the equation to get:
2 * S5 / 5 - A1 = A5
let S5 equal 100 to get:
2 * 100 / 5 - A1 = A5
simplify to get:
40 - A1 = A5
since An = A1 + (n-1)*d, then:
40 - A1 = A1 + (n-1)*d
subtract A1 from both sides of this equation to get:
400 - 2A1 = (n-1)*d
distribute the multiplication to get:
40 - 2A1 = (n-1)*d
since n = 5, then:
40 - 2A1 = 4*d
solve for A1 to get:
A1 = 40 - 4*d)/2
simplify to get:
A1 = 20 - 2d
if this formula is good, then all we need to do is find a value for d that will allow us to subtract from 20 and get an integer (positive or negative should work). that value will be A1.
let's see if this works.
if d = 1, then A1 = 20 - 2 = 18.
A5 = A1 + 4d = 18 + 4 * 1 = 22
S5 = 1/2 * 5 * (18 + 22) = 1/2 * 5 * 40 = 1/2 * 200 = 100
it works !!!!!
if d = 2, then A1 = 20 - 4 = 16
A5 = 16 + 4*2 = 16 + 8 = 24
S5 = 1/2 * 5 * (16 + 24) = 1/2 * 5 * 40 = 1/2 * 200 = 100
id d = 4, then A1 = 20 - 8 = 12
A5 = 12 + 4*4 = 12 + 16 = 28
S5 = 1/2 * 5 * (12 + 28) = 1/2 * 5 * 40 = 1/2 * 200 = 100
it worked for all three !!!!
i used 3 different arithmetic series.
it was a matter of applying the formulas that come with arithmetic series.
An = A1 + (n-1)*d
Sn = n/2 * (A1 + An)
just to see if it will work with a negative number:
start with A1 = 20 - 2d.
let d = 100.
2d = 200
20 - 200 = -180
A1 = -180
d = 100
A5 = -180 + 4*100 = -180 + 400 = 220.
S5 = 5/2 * (-180 + 220) = 5/2 * 5*40/2 = 5*20 = 100
it works again !!!!!
the sequence is -180, -80, 20, 120, 220
add them up and you get 100.
you can probably do the same this with geometric series, and if you made up your own series rules, you might even be able to develop a formula using that.
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