Arithmetic Progressions

**pollardrho06** · May 29th 2010, 02:59 PM

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.

**SpringFan25** · May 29th 2010, 04:16 PM

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)

**pollardrho06** · May 29th 2010, 04:34 PM

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.

**Archie Meade** · May 29th 2010, 04:47 PM

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

$7d=21$

**pollardrho06** · May 29th 2010, 04:57 PM

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!!

**bjhopper** · May 29th 2010, 06:35 PM

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

**Archie Meade** · May 30th 2010, 12:45 AM

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.

**Wilmer** · May 31st 2010, 03:30 AM

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

**Archie Meade** · May 31st 2010, 06:21 AM

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".

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