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In a sequence of 12 numbers, each term, except for the first [#permalink]

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03 Nov 2012, 19:36

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In a sequence of 12 numbers, each term, except for the first one, is 12^11 less than the previous term. If the greatest term in the sequence is 12^12, what is the smallest term in the sequence?

Remember the formula for an arithmetic sequence: An = A1+(n-1)d

Thus, A12 = A1+(12-1)d

Where d is the difference between each pairs of consecutive terms. Since each term is 1211 less than the previous term, d=-1211Click individual answer choices for specific explanations.

From what I undersltand the formula for an arithmetic sequence represents

"some-number-in-the-sequence" = first-number-in-sequence + (the-place-of-that-number - 1) x the-difference-of-each-#-in-the-sequence

I'm not exactly sure how that formula fits into the question, could someone help me to understand?

Re: In a sequence of 12 numbers, each term, except for the first [#permalink]

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03 Nov 2012, 22:24

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anon1 wrote:

In a sequence of 12 numbers, each term, except for the first one, is 12^11 less than the previous term. If the greatest term in the sequence is 12^12, what is the smallest term in the sequence?

−12^11 0 12^11 11·12^11 12^12

In this question, there are total 12 terms. and "greatest" term is 12^12. Each term is 12^11 less than previous one. What does this signify? This shows it is an AP - a decreasing one. (Notice "Each term is 12^11 less than previous one')

Therefore, 12^12 is the first term of the AP and -12^11 is the difference between successive terms, the smallest term should be the last one.

Last term = First term + (n-1)*d therefore, Last term = 12^12 + 11*(-12^11) => Last term = 12^12 - 11*12^11 => Last term = 12^11 (12-11) = 12^11 Which is the smallest term.

Re: In a sequence of 12 numbers, each term, except for the first [#permalink]

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05 Nov 2012, 05:00

bhavinshah5685 wrote:

hi, I am not able to understand one basic...

How can we infer that 12^12 is the first term or the last term in the sequence given in question?? Is it trial and error...???

Thanks in advance..

Question says:

1)each term, except for the first one, is 12^11 less than the previous term => decreasing AP 2)greatest term in the sequence is 12^12 Which term is greatest in a decreasing AP? First one!

Re: In a sequence of 12 numbers, each term, except for the first [#permalink]

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05 Nov 2012, 16:19

I am confused.... "except for the first one," what does this phrase in the question has to say..... remaining sequence is fine... but they never mentioned this series to be a AP series.... what if first term is 12^9? this will be the lowest term right?

Re: In a sequence of 12 numbers, each term, except for the first [#permalink]

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05 Nov 2012, 16:55

In a sequence of 12 numbers, each term, except for the first one, is 12^11 less than the previous term. If the greatest term in the sequence is 12^12, what is the smallest term in the sequence?

The above are the 12 numbers represented by underscores, and the numbers beneath represent their position in the sequence.

The question only mentions "except for the first one" because it wants to state again that there is nothing before the first number. It is a little redundant but the question wants to make itself clear.

The first number can't be 12^9 because it tells you that the numbers get smaller and 12^12 is the biggest number. Therefore 12^12 is the first number.

Re: In a sequence of 12 numbers, each term, except for the first [#permalink]

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06 Nov 2012, 00:33

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anon1 wrote:

Ohhhhh, I see.

I didn't realize that APs could decreasing...

I looked at the question as if 12^12 was the last number in the sequence and that the first (the smallest) was unknown. Thank you as always VIPS

Just a heads up... Although an A.P could be a decreasing one, the one in the question is not. The last term is 12^12 and every term before its is lesser than 12^12. So the first term is 12^11 and last term is 12^12. So the A.P is in fact an increasing one.

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Re: In a sequence of 12 numbers, each term, except for the first [#permalink]

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06 Nov 2012, 01:12

MacFauz wrote:

Just a heads up... Although an A.P could be a decreasing one, the one in the question is not. The last term is 12^12 and every term before its is lesser than 12^12. So the first term is 12^11 and last term is 12^12. So the A.P is in fact an increasing one.

Kudos Please... If my post helped.

And how do you exactly interpret this?

question clearly says: "each term is 12^11 less than the previous term" . This could be true "ONLY IF" its a decreasing AP. Unless you are counting from right to left?

for example in 10, 8, 6, 4... each term is 2 less than the previous term _________________

Re: In a sequence of 12 numbers, each term, except for the first [#permalink]

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06 Nov 2012, 02:00

Haha... Yes.. I was considering the A.P as 12^11, 12^11+12^11,12^11+12^11+12^11 and so on with ther 12th term being 12^11 + 11*12^12... .. I could be wrong though... _________________

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Re: In a sequence of 12 numbers, each term, except for the first [#permalink]

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24 Aug 2013, 00:40

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In a sequence of 12 numbers, each term, except for the first one, is 12^11 less than the previous term. If the greatest term in the sequence is 12^12, what is the smallest term in the sequence?

A. −12^11 B. 0 C. 12^11 D. 11·12^11 E. 12^12

common Difference= 12^11 Last term = 12^12 Last term = First Term + last Term X common difference. 12^11= First term + 12^12 X 12^11 First term=12^11 - 12^12 X 12^11 First term=12^11(12-11) First term=12^11 OA-C

Re: In a sequence of 12 numbers, each term, except for the first [#permalink]

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24 Sep 2013, 17:48

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Points to note in this problem: 1) We have 12 terms 2) Greatest value is 12^12 3) Every other value decreases by 12^11 If we can notice that 12 * 12^11 = 12^12, we can conclude that the smallest value must be 12^11 without doing any math.

Here's the underlying concept: 3^5 + 3^5 + 3^5 = 3*3^5 = 3^6 OR 3^6 - 3^5 - 3^5 = 3^5. --> So if this were a set of "3" terms, it could be {3(3^5), 2(3^5), 3^5} OR {3^6, 2*3^5, 3^5}

Therefore, in this case, 12^11 + 12^11 +...(12 times) = 12*12^11 = 12^12. OR 12^12 - 11 ( 12^11) = 12^11. So the smallest value must be 12^11.

Re: In a sequence of 12 numbers, each term, except for the first [#permalink]

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22 Jun 2015, 13:09

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Re: In a sequence of 12 numbers, each term, except for the first [#permalink]

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22 Jun 2015, 13:21

Vips0000 wrote:

anon1 wrote:

In a sequence of 12 numbers, each term, except for the first one, is 12^11 less than the previous term. If the greatest term in the sequence is 12^12, what is the smallest term in the sequence?

−12^11 0 12^11 11·12^11 12^12

In this question, there are total 12 terms. and "greatest" term is 12^12. Each term is 12^11 less than previous one. What does this signify? This shows it is an AP - a decreasing one. (Notice "Each term is 12^11 less than previous one')

Therefore, 12^12 is the first term of the AP and -12^11 is the difference between successive terms, the smallest term should be the last one.

Last term = First term + (n-1)*d therefore, Last term = 12^12 + 11*(-12^11) => Last term = 12^12 - 11*12^11 => Last term = 12^11 (12-11) = 12^11 Which is the smallest term.

Hence Ans C it is.

Please anyone help me how to get from red to blue in the equality above. Thanks! _________________

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In a sequence of 12 numbers, each term, except for the first one, is 12^11 less than the previous term. If the greatest term in the sequence is 12^12, what is the smallest term in the sequence?

−12^11 0 12^11 11·12^11 12^12

In this question, there are total 12 terms. and "greatest" term is 12^12. Each term is 12^11 less than previous one. What does this signify? This shows it is an AP - a decreasing one. (Notice "Each term is 12^11 less than previous one')

Therefore, 12^12 is the first term of the AP and -12^11 is the difference between successive terms, the smallest term should be the last one.

Last term = First term + (n-1)*d therefore, Last term = 12^12 + 11*(-12^11) => Last term = 12^12 - 11*12^11 => Last term = 12^11 (12-11) = 12^11 Which is the smallest term.

Hence Ans C it is.

Please anyone help me how to get from red to blue in the equality above. Thanks!

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