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Re: Four terms in Arithmetic progression have a sum of 28. [#permalink]
3
Seems difficult at first, but requires good concept knowledge :

a+a+x+a+2x+a+3x = 28
4a+6x = 28
2a+3x = 14

Possibility is out of a,b,c,d two can be negative or all 4 can be positive, so lets test for positive first :
2a+3x = 14
2(2)+3(4) = 14 , a=2 b=6 c=10 d=14 , this doesn't satisfy 5/6 eq.
2(4)+3(2) = 14 , a=4 b=6 c=8 d=10 , this satisfy 5/6 eq.

So d=10

Hope this correct
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Re: Four terms in Arithmetic progression have a sum of 28. [#permalink]
KarunMendiratta wrote:
Explanation:

Let the terms be (x - 3), (x - 1), (x + 1), and (x + 3)
So, (x - 3) + (x- 1) + (x + 1) + (x + 3) = 28
x = 7

Therfore, terms are 4, 6, 8, and 10

Let us check if x = 7 satisfies the second scenario;
\(\frac{(4)(10)}{(6)(8)} = \frac{40}{48} = \frac{5}{6}\)
We have a match

Col. A: 10
Col. B: 11

Hence, option B


How did you arrived at the decision of being the terms equal to (x - 3), (x - 1), (x + 1), and (x + 3)?
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Re: Four terms in Arithmetic progression have a sum of 28. [#permalink]
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koala wrote:
KarunMendiratta wrote:
Explanation:

Let the terms be (x - 3), (x - 1), (x + 1), and (x + 3)
So, (x - 3) + (x- 1) + (x + 1) + (x + 3) = 28
x = 7

Therfore, terms are 4, 6, 8, and 10

Let us check if x = 7 satisfies the second scenario;
\(\frac{(4)(10)}{(6)(8)} = \frac{40}{48} = \frac{5}{6}\)
We have a match

Col. A: 10
Col. B: 11

Hence, option B


How did you arrived at the decision of being the terms equal to (x - 3), (x - 1), (x + 1), and (x + 3)?


While solving the Qs from AP and/or GP, assume the consecutive numbers as ... (a - 3d), (a - 2d), (a- d), a, (a + d), (a + 2d), (a + 3d), ....
This will help you save a lot of time.
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Re: Four terms in Arithmetic progression have a sum of 28. [#permalink]
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