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# Four terms in arithmetic progression have sum of 28. Product

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Four terms in arithmetic progression have sum of 28. Product  [#permalink]

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Updated on: 05 Jan 2014, 05:50
17
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Difficulty:

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Question Stats:

65% (02:29) correct 35% (02:47) wrong based on 207 sessions

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Four terms in arithmetic progression have sum of 28. Product of the extreme terms and that of the middle terms are in the ratio 5:6. Find the largest term?

A. 8
B. 9
C. 10
D. 11
E. 12

Originally posted by jadixit on 05 Jan 2014, 03:32.
Last edited by Bunuel on 05 Jan 2014, 05:50, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Four terms in arithmetic progression have sum of 28. Product  [#permalink]

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28 Sep 2015, 02:16
4
1
Four terms in arithmetic progression have sum of 28. Product of the extreme terms and that of the middle terms are in the ratio 5:6. Find the largest term?

A. 8
B. 9
C. 10
D. 11
E. 12

Using brute force can be quite helpful.

Sum of 28 means that essentially all 4 numbers are equal to 7 each. Since the product of 1st and 4th is different from the product of 2nd and 3rd terms, the common difference must be more than 0.
So the numbers could be 4, 6, 8, 10 (with 7 as the mean) when the common difference is more than 0 but least possible (assuming integers)
Product of 1st and 4th = 40
Product of 2nd and 3rd = 48
Ratio = 5:6 (what we wanted)
So the largest term is 10.

Had we not obtained the required ratio, we would have increased the common difference and taken the terms as 1, 5, 9, 11 and then checked. This would have given us a pattern to tell us whether the ratio is increasing or decreasing.
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Re: Arithmetic progression  [#permalink]

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05 Jan 2014, 03:45
5
3
Four terms in arithmetic progression have sum of 28. Product of the extreme terms and that of the middle terms are in the ratio 5:6. Find the largest term?

A 8
B 9
C 10
D 11
E 12

lets take the terms as (a-3d), (a-d), (a+d), (a+3d).
as sum = 28, so a = 7

Lets check the option and we can see 7+3d can be equal to 10. lets test that scenario.

So the numbers can be 4,6,8,10. product of extreme terms/product of middle terms = 40/48 = 5/6 So Option C)

We can use this technique to solve GMAT problems as none of the above or cannot be determined is yet to be an option in GMAT.

For the Maths purists :

(a-3d) * (a+3d) /(a-d) * (a+d) = 5/6 => 6(a^2 - 9d^2) = 5(a^2 - d^2) => a^2 = 49d^2 as a = 7 then d has to be 1. hence option c)

As we don't get marks for steps I prefer the approach 1.
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##### General Discussion
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Re: Four terms in arithmetic progression have sum of 28. Product  [#permalink]

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04 Jul 2014, 08:34
kinjiGC wrote:
Four terms in arithmetic progression have sum of 28. Product of the extreme terms and that of the middle terms are in the ratio 5:6. Find the largest term?

A 8
B 9
C 10
D 11
E 12

lets take the terms as (a-3d), (a-d), (a+d), (a+3d).
as sum = 28, so a = 7

Lets check the option and we can see 7+3d can be equal to 10. lets test that scenario.

So the numbers can be 4,6,8,10. product of extreme terms/product of middle terms = 40/48 = 5/6 So Option C)

We can use this technique to solve GMAT problems as none of the above or cannot be determined is yet to be an option in GMAT.

For the Maths purists :

(a-3d) * (a+3d) /(a-d) * (a+d) = 5/6 => 6(a^2 - 9d^2) = 5(a^2 - d^2) => a^2 = 49d^2 as a = 7 then d has to be 1. hence option c)

As we don't get marks for steps I prefer the approach 1.

Lets check the option and we can see 7+3d can be equal to 10. lets test that scenario.

So the numbers can be 4,6,8,10. product of extreme terms/product of middle terms = 40/48 = 5/6 So Option C)

Question : how did you know 7 + 3d can be equal to 10?

How did you get #'s 4,6, 8, 10?
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Re: Four terms in arithmetic progression have sum of 28. Product  [#permalink]

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04 Jul 2014, 09:04
sagnik242 wrote:
kinjiGC wrote:
Four terms in arithmetic progression have sum of 28. Product of the extreme terms and that of the middle terms are in the ratio 5:6. Find the largest term?

A 8
B 9
C 10
D 11
E 12

lets take the terms as (a-3d), (a-d), (a+d), (a+3d).
as sum = 28, so a = 7

Lets check the option and we can see 7+3d can be equal to 10. lets test that scenario.

So the numbers can be 4,6,8,10. product of extreme terms/product of middle terms = 40/48 = 5/6 So Option C)

We can use this technique to solve GMAT problems as none of the above or cannot be determined is yet to be an option in GMAT.

For the Maths purists :

(a-3d) * (a+3d) /(a-d) * (a+d) = 5/6 => 6(a^2 - 9d^2) = 5(a^2 - d^2) => a^2 = 49d^2 as a = 7 then d has to be 1. hence option c)

As we don't get marks for steps I prefer the approach 1.

Lets check the option and we can see 7+3d can be equal to 10. lets test that scenario.

So the numbers can be 4,6,8,10. product of extreme terms/product of middle terms = 40/48 = 5/6 So Option C)

Question : how did you know 7 + 3d can be equal to 10?

How did you get #'s 4,6, 8, 10?

When using the options , start with option C). Normally the options will be ordered (How I know it, I also develop question items )
if option C) doesn't satisfy the equation, at least you can have a range of numbers which might satisfy and you can eliminate the options easily.

Coming to the question:
if 7+3d = 10, so we can calculate the value of d= 1 and we can calculate the other three numbers.
as 1st term is a-3d, so putting a =7 and d =1, we can get 4 and so on.

Any doubts, please revert.
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Re: Four terms in arithmetic progression have sum of 28. Product  [#permalink]

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04 Jul 2014, 15:59
Why did you set up the equations (a-3d), (a-d), ect
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Re: Four terms in arithmetic progression have sum of 28. Product  [#permalink]

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05 Jul 2014, 01:15
3
4
bankerboy30 wrote:
Why did you set up the equations (a-3d), (a-d), ect

That is simply as assumption. It reduces the calculation time and complexity.

If we have three terms in AP, we can assume
a-d, a, a+d

If we have four terms in AP, we can assume
a-3d, a-d, a+d, a+3d

if we have 5 terms in AP, we can assume
a-2d, a-d, a, a+d, a+2d

So if as we add the terms, the terms cancel each other and we can have some advantage in calculation.
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Four terms in arithmetic progression have sum of 28. Product  [#permalink]

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Updated on: 14 Sep 2019, 10:39
Four terms in arithmetic progression have sum of 28. Product of the extreme terms and that of the middle terms are in the ratio 5:6. Find the largest term?

A. 8
B. 9
C. 10
D. 11
E. 12

let a=1st term and d=difference between terms
4a+6d=28→2a+3d=14
2 possibilities: a=1; d=4 and a=4; d=2
but only 4 and 2 give 5:6 extreme:middle term product ratio
so 10
C

Originally posted by gracie on 01 Jan 2017, 20:03.
Last edited by gracie on 14 Sep 2019, 10:39, edited 2 times in total.
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Re: Four terms in arithmetic progression have sum of 28. Product  [#permalink]

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16 May 2019, 06:02
30 seconds method

W+X+Y+Z = 28

To get X and Y

Mean: 28/4 = 7
Average of 7: (6+8)/2

X= 6 Y=7 (closest to the mean)

To get W,Z

Given, Ratio 5:6
WZ:48 = 5:6
WZ= 40

W= 4 ; Z= 10
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Re: Four terms in arithmetic progression have sum of 28. Product  [#permalink]

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16 May 2019, 19:02
I have used the following steps:

1 - Set $$A, A+n, A+2n, A+3n$$, SUM OF SET is $$4A+6n = 28$$
2 - Simplify to $$2A+3n = 14$$
3 - $$A + A + 3n = 14$$ and thus $$A + 3n = 14 - A$$ (Meaning the largest is 14 minus smallest)
4 - Started picking numbers from C (POE strategy), $$A + 3n = 10$$ (Largest is 10), $$14 - A = 10$$ thus A should be 4.
5 - $$4 + 3n = 10$$ >>> $$n = 2$$
6 - $$\frac{(10 * 4)}{(6 * 8)}$$ = $$\frac{40}{48}$$ = $$\frac{5}{6}$$

May take more than a minute to solve, however. But I hope it could still help to understand the logic behind. Please correct me if I am wrong!
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Re: Four terms in arithmetic progression have sum of 28. Product  [#permalink]

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14 Sep 2019, 09:10
1
Four terms in arithmetic progression have sum of 28. Product of the extreme terms and that of the middle terms are in the ratio 5:6. Find the largest term?

A. 8
B. 9
C. 10
D. 11
E. 12

Well, I was trying to solve the numbers but then realised something. If the product of extreme terms is in the form of 5x that means one of them has to be a 5 or a multiple of 5.
The only option is C which is 10.
Is that fair?

Bunuel
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Re: Four terms in arithmetic progression have sum of 28. Product  [#permalink]

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14 Sep 2019, 09:19
TheNightKing wrote:
Four terms in arithmetic progression have sum of 28. Product of the extreme terms and that of the middle terms are in the ratio 5:6. Find the largest term?

A. 8
B. 9
C. 10
D. 11
E. 12

Well, I was trying to solve the numbers but then realised something. If the product of extreme terms is in the form of 5x that means one of them has to be a 5 or a multiple of 5.
The only option is C which is 10.
Is that fair?

Bunuel

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Seems fair to me.
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Four terms in arithmetic progression have sum of 28. Product  [#permalink]

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14 Sep 2019, 09:20
Quote:
Bunuel
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Seems fair to me.

Thank you!
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Re: Four terms in arithmetic progression have sum of 28. Product  [#permalink]

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14 Sep 2019, 10:49
Four terms in arithmetic progression have sum of 28. Product of the extreme terms and that of the middle terms are in the ratio 5:6. Find the largest term?

A. 8
B. 9
C. 10
D. 11
E. 12

the AP of the given series of no would be of possiblity 4,6,8,10
4*10/6*8 = 5:6
IMO C ; 10
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Re: Four terms in arithmetic progression have sum of 28. Product  [#permalink]

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14 Sep 2019, 11:01
Four terms in arithmetic progression have sum of 28. Product of the extreme terms and that of the middle terms are in the ratio 5:6. Find the largest term?

A. 8
B. 9
C. 10
D. 11
E. 12

Given:
1. Four terms in arithmetic progression have sum of 28.
2. Product of the extreme terms and that of the middle terms are in the ratio 5:6.

Asked: Find the largest term?

Let the 4 terms in AP be {a-3d, a-d , a+d, a+3d}
4a = 28; a = 7
$$\frac{(7-3d)(7+3d)}{(7-d)(7+d)} = \frac{5}{6}$$
$$\frac{49 - 9d^2}{49 - d^2} = \frac{5}{6}$$
$$\frac{49 - 9d^2}{49 - d^2} = \frac{5}{6}$$
6(49 - 9d^2) = 5(49-d^2)
49 = 54d^2 - 5d^2 = 49d^2
d = +-1

Terms = {4,6,8,10}
Largest term = 10

IMO C
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Re: Four terms in arithmetic progression have sum of 28. Product   [#permalink] 14 Sep 2019, 11:01
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