1. ## Arithmetic Progressions

Hi all.

I'm working on the following problem:

An arithmetic progression has 14 terms. The sum of the odd-numbered terms is 140 and the sum of the even-numbered terms is 161. Find the common difference of the progression and the 14th term.

Here's what I've done with it so far...

A.P.

n = 14

T_1 + T_3 + T_5 + . . . + T_13 = 140

T_2 + T_4 + T_6 + . . . + T_14 = 161

T_1 + T_2 + . . . + T_14 = 140 + 161 = 301 = S_14

Now, for an A.P.,

S_n = n / 2 [ 2 * a + (n - 1) * d ]

S_14 = 14 / 2 [ 2 * a + (14 - 1) * d ] = 301

7 [ 2a + 13d ] = 301

2a + 13d = 43 . . . . . . (i)

So, I get one equation... Now, I need another equation in terms of a and d so that I can solve them simultaneously to find a and d and then the remaining question will be done...

Any help will be greatly appreciated.

Thanks,

Shahz.

2. 7 odd terms: $S_{odd} = x_1 + x_3 + ...x_{13} = 140$
7 even terms: $S_{even} = x_2 + x_4 + ...x_{14} = 161$

but consider:
$S_{even} - S_{odd} = (x_2 + x_4 + ...x_{14}) - (x_1 + x_3 + ...x_{13} ) = 161 - 140$

$S_{even} - S_{odd} = (x_2 + x_4 + ...x_{14}) - (x_1 + x_3 + ...x_{13} ) = 21$

$S_{even} - S_{odd} = (x_2-x_1) + (x_4-x_3) .....+(x_{14}-x_{13}) = 21$
each bracketed expression is teh difference between two consecutive terms of the AP. There are 7 of them in total.

But you KNOW for an AP that the difference between consectuve terms is just d (the common difference)

$S_{even} - S_{odd} = d + d +.... +d = 21$
$S_{even} - S_{odd} = 7d = 21$
That gives you d directly.

Then , you can use your equation to work out $a$ and then get $x_{14}$ (assuming your equation is right, i didn't check it)

3. Originally Posted by SpringFan25
There are easier ways to do this:

7 odd terms: $S_{odd} = x_1 + x_3 + ...x_{13} = 140$
7 even terms: $S_{even} = x_2 + x_4 + ...x_{14} = 161$

but consider:
$S_{even} - S_{odd} = (x_2 + x_4 + ...x_{14}) - (x_1 + x_3 + ...x_{13} ) = 161 - 140$

$S_{even} - S_{odd} = (x_2 + x_4 + ...x_{14}) - (x_1 + x_3 + ...x_{13} ) = 21$

$S_{even} - S_{odd} = (x_2-x_1) + (x_4-x_3) .....+(x_{14}-x_{13}) = 21$
each bracketed expression is teh difference between two consecutive terms of the AP. There are 7 of them in total.

But you KNOW for an AP that the difference between consectuve terms is just d (the common difference)

$S_{even} - S_{odd} = d + d +.... +d = 21$
$S_{even} - S_{odd} = 7d = 21$
That gives you d directly.

Then , you can use your equation to work out $x_{14}$ (assuming your equation is right, i didn't check it)
Wonderfully elegant solution!! Thank you so much.

4. Originally Posted by pollardrho06
Hi all.

I'm working on the following problem:

An arithmetic progression has 14 terms. The sum of the odd-numbered terms is 140 and the sum of the even-numbered terms is 161. Find the common difference of the progression and the 14th term.

Here's what I've done with it so far...

A.P.

n = 14

T_1 + T_3 + T_5 + . . . + T_13 = 140

T_2 + T_4 + T_6 + . . . + T_14 = 161

T_1 + T_2 + . . . + T_14 = 140 + 161 = 301 = S_14

Now, for an A.P.,

S_n = n / 2 [ 2 * a + (n - 1) * d ]

S_14 = 14 / 2 [ 2 * a + (14 - 1) * d ] = 301

7 [ 2a + 13d ] = 301

2a + 13d = 43 . . . . . . (i)

So, I get one equation... Now, I need another equation in terms of a and d so that I can solve them simultaneously to find a and d and then the remaining question will be done...

Any help will be greatly appreciated.

Thanks,

Shahz.
Alternatively....

the odd terms are

$a,\ a+2d,\ a+4d,\ ......$ 7 terms, 6 multiples of "d"

the even terms are

$a+d,\ a+3d,\ a+5d,......$ 7 terms, 7 multiples of "d"

then

$7a+(2+4+6+8+10+12)d=140$

$7a+42d=140$

and

$7a+(1+3+5+7+9+11+13)d$

$7a+49d=161$

the difference is 7d

$7d=21$

5. Originally Posted by Archie Meade
Alternatively....

the odd terms are

$a,\ a+2d,\ a+4d,\ ......$ 7 terms, 6 multiples of "d"

the even terms are

$a+d,\ a+3d,\ a+5d,......$ 7 terms, 7 multiples of "d"

then

$7a+(2+4+6+8+10+12)d=140$

$7a+42d=140$

and

$7a+(1+3+5+7+9+11+13)d$

$7a+49d=161$

the difference is 7d

$7d=21$
Fantastic!! Thanks!!

6. ## arithmetric progression

Originally Posted by pollardrho06
Fantastic!! Thanks!!
Hello shahz,

I did this differently

sum of 14 terms = 140+161=301
using the sum formula it simplifies to 2a +13d=43
I tried a=1 no go a=2 go 13d=39 d=3

progression 2,5,8,11,........

The sum of the odds is 161 not 140 and the sum of the evens is 140 not 161.

bjh

7. Originally Posted by bjhopper
Hello shahz,

I did this differently

sum of 14 terms = 140+161=301
using the sum formula it simplifies to 2a +13d=43
I tried a=1 no go a=2 go 13d=39 d=3

progression 2,5,8,11,........

The sum of the odds is 161 not 140 and the sum of the evens is 140 not 161.

bjh
The "odds" are the odd-numbered terms T1, T3, T5 etc
and the "evens" are the even-numbered terms T2, T4, T6 etc

not the odd and even values in the sequence itself.

8. Originally Posted by pollardrho06
An arithmetic progression has 14 terms. The sum of the odd-numbered terms is 140 and the sum of the even-numbered terms is 161. Find the common difference of the progression and the 14th term.
Anything wrong with this "short-cut"? :

odd: 140 / 7 = 20 (4th odd# term)
even:161 / 7 = 23 (4th even# term)
So common diff = 23 - 20 = 3 (or 6 individually)
And 14th term = 7th even term = 23 + 3(6) = 41

9. Originally Posted by Wilmer
Anything wrong with this "short-cut"? :

odd: 140 / 7 = 20 (4th odd# term)
even:161 / 7 = 23 (4th even# term)
So common diff = 23 - 20 = 3 (or 6 individually)
And 14th term = 7th even term = 23 + 3(6) = 41
Nothing wrong with it at all!

Odd numbered terms sum as follows...

$a+(a+2d)+(a+4d)+(a+6d)+(a+8d)+(a+10d)+(a+12d)$

$=(a+6d-6d)+(a+6d-4d)+(a+6d-2d)+$ $(a+6d)+(a+6d+2d)+(a+6d+4d)+(a+6d+6d)$

$=(a+6d)+(a+6d)+(a+6d)+(a+6d)$ $+(a+6d)+(a+6d)+(a+6d)+6d-6d+4d-4d+2d-2d$

$=7(a+6d)=7(middle\ term)=140$

$middle\ term=T_7=a+6d=\frac{140}{7}=20$

Doing the same thing with the even numbered terms discovers "d" and "a".