An arithmetic progression problem

[Anzas]

in an arithmetic sequence there is an even number of term's
the sum of terms in the odd places is 440 and the sum of terms in the even places is 520, the last term is bigger than the first term by 156
find how many term's the arithmetic sequence has.

 

[matt grime]

perhaps you should post this in the homework section?

 

[poolwin2001]

I Suppose you know the formula of summation of series:
$$S=\frac{n}{2}(2a + (n-1)d)$$
(If not it is easy to derive)
where $$a$$is the first term $$n$$ is the number of terms and
$$d$$ is the common difference between them

Hint:
Let the series be:
$$a,a+d,a+2d,...$$

even terms $$a,a+2d,...$$
odd terms $$a+d,a+3d,...$$

These are Sequences with common diference 2d.
Use their sum to get 2 eqns
Using last term - first term = 156 you have 3 eqns 3 unknowns(a,n,d).
(Need anymore help)
P.S:Give me Homework Helper medal Quickly PLEEEEEASE!
EDIT:I gave the wrong formula as I was entranced by the latex graphics
nobody but halls of ivy saw it i think
EDIT:Derivation of formula
$$a,a+d,a+2d,...,a+(n-1)d (i)$$
invert the above
$$a+(n-1)d,a+(n-2)d,a+2d,...,a+d,a (ii)$$
$$(i)+(ii)=> 2S=2na + n(n-1)d ==> s=\frac{n}{2}(2a + (n-1)d)$$
This was proposed by gauss(I think)

 

[HallsofIvy]

Since this is an arithmetic sequence then, taking a₁ as the first term in the sequence, a₂= a₁+ n, a₃= a₂+ n= a₁+ 2n and, in general, a_i= a₀+ (i-1)n.

If there are N numbers in the sequence then the last number is a_N= a₁+ (N-1)n and so the difference between the first and last terms is
a_N-₁= (N-1)n= 156.
The sum of the even terms is a₂+a₄+a₆+...a_N=(a₁+n)+(a₁+ 3n)+ (a₁+ 5n)+ ...+ (a₁+ (N-1)n= (N/2)a₁+n(1+ 3+ 5+ ...+ (N-1)).

Now, find a formula for 1+ 3+ 5+ ...+ N-1 so you can get another equation for n and N.

 

[Anzas]

got it solved thanks for your help

 

[needhelpperson]

I tried out this problem, but i can't seem to go ne where with it.

For sum of even numbers

(a+d+a+2d(n-1))*n = 1040<------------ 2an + 2dn^2 -dn = 1040

For sum of odd numbers

(a + a +2d(n-1))*n = 880<-------------- 2an + 2dn^2 - 2dn = 880

solved the two system of equation

dn = 160

N = total number = 2n

Nd = 320

d(N-1) = 156<----------- d = 164

so solve for N using Nd = 320 = 320/164 = 1.95...

Obviously this is not correct. What did i do wrong here?

 

[Anzas]

here is my solution i hope its understandable enough

formula's:
aN = a1 + dn - d
sN = (2a1 + d(n - 1))*n/2

odd:
s = 440
d = 2d
n = n
aN = aN - d
a1 = a1

even:
s = 520
d = 2d
n = n
aN = aN
a1 = a1 + d

general:
s = 960
d = d
n = 2n
aN = aN
a1 = a1

*****************
aN - a1 = 156
aN = a1 + 2dn - d
aN - a1 = 2dn - d
156 = 2dn - d
156 = d(2n - 1)
*****************

even:
520 = (2a1 + 2d + 2dn - 2d)*n/2
520 = (2a1 + 2d + d(2n - 2))*n/2
520 = (2a1 + 2d + 2dn - 2d)*n/2
520 = (a1 + d + dn - d)*n
520 = a1n + dn^2
a1n = 520 - dn^2
a1 = (520 - dn^2)/n

general:
960 = (2a1 + 2dn - d)*2n/2
960 = (2a1 + 2dn - d)*n
960 = (2a1 + d(2n - 1)*n
960 = 2a1 + 156n
2a1 = 960 - 156n
a1 = 480 - 78n

odd:
440 = (2a1 + 2dn -2d)*n/2
440 = (a1 + dn - d)*n
440 = a1n + dn^2 - dn
a1n = 440 - dn^2 + dn
a1 = (440 - dn^2 + dn)/n


***************
440 - dn^2 + dn = 520 - dn^2
440 + dn = 520
dn = 80
d = 80/n
***************

***************
156 = d(2n - 1)
***************

156 = 80(2n - 1)/n
156n = 80(2n - 1)
156n = 160n - 80
4n = 80
n = 20
2n = 40

d = 80/n
d = 80/20
d = 4