# Arithmetic progression(alternate method)

1. Sep 6, 2004

### needhelpperson

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.

This was from a previous post, but i wanted to figure it out this way.

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?

2. Sep 6, 2004

### Leong

For the even sequence, you should get this

n[(a+d)+(a+d)+(n-1)2d]=1040

Solve the 2 equations, get nd=80;Nd=160; Nd-d=156; d=4; N=40.

3. Sep 6, 2004

### needhelpperson

thanks alot, so that's where i went wrong.

4. Sep 6, 2004

### Tide

Here's how I did it:

If you multiply the number of terms in an arithmetic series by the average of the first and last terms the product is the sum of the series. Therefore, the sum of the even terms is

$$\frac {N}{2} \frac {a_2+a_N}{2} = 520$$

and the sum of the odd terms is

$$\frac {N}{2} \frac {a_1+a_{N-1}}{2} = 440$$

(Note the sum of the odd terms is even so the $\frac {N}{2}$ is a whole number)

Now, $a_2 = a_1+d$ and $a_{N-1} = a_N-d$ where d is the common difference. Substitute these into the equations and subtract the two equations to obtain $N d = 160$
We also know $(N-1)d = 156$ so, dividing gives

$$\frac {N}{N-1} = \frac{160}{156} = \frac {40}{39}$$

from which N = 40 follows.