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by phoenix9801 » Sat May 26, 2012 3:50 am
Can Someone show me how to solve using plugging in numbers not in algebraically. Please be simple as possible in details. Thanks

Someone who to solve it the most simple to understanding sequence question please because it get difficult and tricky on the GMAT. Thanks



S is sequence S1, S2,S3...Sn in which every term after the first is one less then three times the previous term. If S5 - S3= 28, which of the following is the first term in the sequence?

a) 2/3
b) 8/9
c) 1
d) 5/3
e) 2
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by Anurag@Gurome » Sat May 26, 2012 5:52 am
phoenix9801 wrote:S is sequence S1, S2,S3...Sn in which every term after the first is one less then three times the previous term. If S5 - S3= 28, which of the following is the first term in the sequence?
This is not that kind of problem which can be solved with plugging numbers. Rather plugging the options is an approach but that would require a bit more time in this case depending upon what you are fluent with. Anyway, I'm showing the plugging options approach first and the algebraic one later. You decide yourself which one you want.

Plugging Options Approach:
  • A. Say, S1 = 2/3
    Hence, S2 = 3*(2/3) - 1 = 1
    Hence, S3 = 3*1 - 1 = 2
    Hence, S4 = 3*2 - 1 = 5
    Hence, S5 = 3*5 - 1 = 14

    Hence, S5 - S3 = 14 - 2 = 12 ≠ 28 ---> NOT the correct Option

    B. Say, S1 = 8/9
    Hence, S2 = 3*(8/9) - 1 = 5/3
    Hence, S3 = 3*(5/3) - 1 = 4
    Hence, S4 = 3*4 - 1 = 11
    Hence, S5 = 3*11 - 1 = 32

    Hence, S5 - S3 = 32 - 4 = 28 ---> CORRECT Option.
Algebraic Approach:
Say, the first term is S1 = x.
Hence, S2 = (3x - 1)
Hence, S3 = 3(3x - 1) - 1 = (9x - 4)
Hence, S4 = 3(9x - 4) - 1 = (27x - 13)
Hence, S5 = 3(27x - 13) - 1 = (81x - 40)

Hence, S5 - S3 = (81x - 40) - (9x - 4) = (72x - 36) = 28

Hence, x = (28 + 36)/72 = 64/72 = 8/9

The correct answer is B.
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by phoenix9801 » Sat May 26, 2012 10:53 am
Thank you so much for your help
Anurag@Gurome wrote:
phoenix9801 wrote:S is sequence S1, S2,S3...Sn in which every term after the first is one less then three times the previous term. If S5 - S3= 28, which of the following is the first term in the sequence?
This is not that kind of problem which can be solved with plugging numbers. Rather plugging the options is an approach but that would require a bit more time in this case depending upon what you are fluent with. Anyway, I'm showing the plugging options approach first and the algebraic one later. You decide yourself which one you want.

Plugging Options Approach:
  • A. Say, S1 = 2/3
    Hence, S2 = 3*(2/3) - 1 = 1
    Hence, S3 = 3*1 - 1 = 2
    Hence, S4 = 3*2 - 1 = 5
    Hence, S5 = 3*5 - 1 = 14

    Hence, S5 - S3 = 14 - 2 = 12 ≠ 28 ---> NOT the correct Option

    B. Say, S1 = 8/9
    Hence, S2 = 3*(8/9) - 1 = 5/3
    Hence, S3 = 3*(5/3) - 1 = 4
    Hence, S4 = 3*4 - 1 = 11
    Hence, S5 = 3*11 - 1 = 32

    Hence, S5 - S3 = 32 - 4 = 28 ---> CORRECT Option.
Algebraic Approach:
Say, the first term is S1 = x.
Hence, S2 = (3x - 1)
Hence, S3 = 3(3x - 1) - 1 = (9x - 4)
Hence, S4 = 3(9x - 4) - 1 = (27x - 13)
Hence, S5 = 3(27x - 13) - 1 = (81x - 40)

Hence, S5 - S3 = (81x - 40) - (9x - 4) = (72x - 36) = 28

Hence, x = (28 + 36)/72 = 64/72 = 8/9

The correct answer is B.
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by phoenix9801 » Sat May 26, 2012 10:10 pm
Question when use the plugging in number method and with a sequence question like this. I read the problem more then one time try to figure it out. How did you know that you supposed to take each answer and then plug it in the next equation.?????????

B. Say, S1 = 8/9
Hence, S2 = 3*(8/9) - 1 = 5/3
Hence, S3 = 3*(5/3) - 1 = 4
Hence, S4 = 3*4 - 1 = 11
Hence, S5 = 3*11 - 1 = 32

Hence, S5 - S3 = 32 - 4 = 28 ---> CORRECT Option.
Anurag@Gurome wrote:
phoenix9801 wrote:S is sequence S1, S2,S3...Sn in which every term after the first is one less then three times the previous term. If S5 - S3= 28, which of the following is the first term in the sequence?
This is not that kind of problem which can be solved with plugging numbers. Rather plugging the options is an approach but that would require a bit more time in this case depending upon what you are fluent with. Anyway, I'm showing the plugging options approach first and the algebraic one later. You decide yourself which one you want.

Plugging Options Approach:
  • A. Say, S1 = 2/3
    Hence, S2 = 3*(2/3) - 1 = 1
    Hence, S3 = 3*1 - 1 = 2
    Hence, S4 = 3*2 - 1 = 5
    Hence, S5 = 3*5 - 1 = 14

    Hence, S5 - S3 = 14 - 2 = 12 ≠ 28 ---> NOT the correct Option

    B. Say, S1 = 8/9
    Hence, S2 = 3*(8/9) - 1 = 5/3
    Hence, S3 = 3*(5/3) - 1 = 4
    Hence, S4 = 3*4 - 1 = 11
    Hence, S5 = 3*11 - 1 = 32

    Hence, S5 - S3 = 32 - 4 = 28 ---> CORRECT Option.
Algebraic Approach:
Say, the first term is S1 = x.
Hence, S2 = (3x - 1)
Hence, S3 = 3(3x - 1) - 1 = (9x - 4)
Hence, S4 = 3(9x - 4) - 1 = (27x - 13)
Hence, S5 = 3(27x - 13) - 1 = (81x - 40)

Hence, S5 - S3 = (81x - 40) - (9x - 4) = (72x - 36) = 28

Hence, x = (28 + 36)/72 = 64/72 = 8/9

The correct answer is B.
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by Anurag@Gurome » Sun May 27, 2012 1:39 am
phoenix9801 wrote:How did you know that you supposed to take each answer and then plug it in the next equation?
When any question gives some properties or information about a variable or unknown sufficient to form one (or more than one) equation(s) with it and asks to find out the value of the variable and it also provides some options for the value including a right one, we can always put the options in the original question to see which one of them satisfies the conditions.

It's essentially a back calculation process.
We can't do the same without any options. That would be finding a needle in a haystack.

Also the same cannot be done for questions without any variable or unknown. For example,
  • How many integers are there between 1 and 100?
    • A. 90
      B. 97
      C. 98
      D. 99
      E. 100
cannot be solved by plugging options as there is no provision for doing so.

But...
  • If x = 2x - 1, what is the value of x?
    • A. -2
      B. -1
      C. 0
      D. 1
      E. 2
can be solved by plugging options just by putting each of the options in place of x in the original equation.

Hope it helps.
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by GMATGuruNY » Sun May 27, 2012 3:09 am
phoenix9801 wrote:Can Someone show me how to solve using plugging in numbers not in algebraically. Please be simple as possible in details. Thanks

Someone who to solve it the most simple to understanding sequence question please because it get difficult and tricky on the GMAT. Thanks



S is sequence S1, S2,S3...Sn in which every term after the first is one less then three times the previous term. If S5 - S3= 28, which of the following is the first term in the sequence?

a) 2/3
b) 8/9
c) 1
d) 5/3
e) 2
We can plug in the answers, which represent the value of the 1st term.
Since integers are easier to manipulate than fractions, we should start with C, an integer value right in the middle.

Answer choice C:
1st term = 1.
2nd term = 3*1 - 1 = 2.
3rd term = 3*2 - 1 = 5.
4th term = 3*5 - 1 = 14.
5th term = 3*14 = 42 - 1 = 41.
5th term - 3rd term = 41-5 = 36.
The difference is too great, implying that the correct answer must be smaller.
Eliminate C, D and E.

Of the two remaining answer choices, 2/3 seems easier to plug in.

Answer choice A:
1st term = 2/3
2nd term = 3(2/3) - 1 = 1.
Since 1 is the 1st term in answer choice C, we calculated above the remaining 3 terms for answer choice A:
2, 5, 14.
5th term - 3rd term = 14-2 = 12.
The difference is too small.
Eliminate A.

The correct answer is B.
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