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What is the sixtieth term in the following sequence? 1, 2, 4

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What is the sixtieth term in the following sequence? 1, 2, 4  [#permalink]

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New post 24 Jul 2010, 11:37
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What is the sixtieth term in the following sequence? 1, 2, 4, 7, 11, 16, 22...

(A) 1,671
(B) 1,760
(C) 1,761
(D) 1,771
(E) 1,821
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Re: Help:Tough problem on exponential sequence  [#permalink]

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New post 24 Jul 2010, 12:01
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praveengmat wrote:
What is the sixtieth term in the following sequence? 1, 2, 4, 7, 11, 16, 22...
(A) 1,671
(B) 1,760
(C) 1,761
(D) 1,771
(E) 1,821


Hi and welcome to Gmat Club. Below is the solution for your question.

First we should find out what is the pattern of this sequence. Write the terms of the sequence as below:

\(a_1=1\);
\(a_2=2=a_1+1\);
\(a_3=4=a_1+1+2\);
\(a_4=7=a_1+1+2+3\);
\(a_5=11=a_1+1+2+3+4\);
\(a_6=16=a_1+1+2+3+4+5\);
\(a_7=22=a_1+1+2+3+4+5+6\);
...

So the \(n_{th}\) term of this sequence equals to first term plus the sum of first \(n-1\) integers.

\(a_{60}=a_1+1+2+3+...+59=a_1+\frac{1+59}{2}*59=1771\).

Answer: D.

Hope it's clear.
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Re: Help:Tough problem on exponential sequence  [#permalink]

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New post 24 Jul 2010, 11:57
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praveengmat wrote:
What is the sixtieth term in the following sequence? 1, 2, 4, 7, 11, 16, 22...
(A) 1,671
(B) 1,760
(C) 1,761
(D) 1,771
(E) 1,821


The difference between consecutive terms of the sequence are in AP
2-1 = 1; 4-2 = 2; 7-4 = 3;

\(T_n = S_n - S_{n-1}\)

\(S_n = 1 + 2 + 4 + 7.......\) n terms \(= 1+ (1 + 1) + (2+2) + (3+4)............\)

\(S_n = 1 + (1+2+3+4 ....\) n-1 terms ) \(+ 1+2+4+7+11......\)n-1 terms

=> \(S_n = 1 + \frac{n*(n-1)}{2} + S_{n-1}\)

=> \(S_n - S_{n-1} = T_n = 1 + \frac{n*(n-1)}{2}\)

\(T_{60} = 1 + \frac{60*(60-1)}{2} = 1+ 30*59 = 1771\)

hence D
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Re: Help:Tough problem on exponential sequence  [#permalink]

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New post 24 Jul 2010, 12:07
gurpreetsingh wrote:
praveengmat wrote:
What is the sixtieth term in the following sequence? 1, 2, 4, 7, 11, 16, 22...
(A) 1,671
(B) 1,760
(C) 1,761
(D) 1,771
(E) 1,821


The difference between consecutive terms of the sequence are in AP
2-1 = 1; 4-2 = 2; 7-4 = 3;

\(T_n = S_n - S_{n-1}\)

\(S_n = 1 + 2 + 4 + 7.......\) n terms \(= 1+ (1 + 1) + (2+2) + (3+4)............\)

\(S_n = 1 + (1+2+3+4 ....\) n-1 terms ) \(+ 1+2+4+7+11......\)n-1 terms

=> \(S_n = 1 + \frac{n*(n-1)}{2} + S_{n-1}\)

=> \(S_n - S_{n-1} = T_n = 1 + \frac{n*(n-1)}{2}\)

\(T_{60} = 1 + \frac{60*(60-1)}{2} = 1+ 30*59 = 1771\)

hence D


\(S_n =\) sum of n term
\(T_n =\) nth term

The reason why I have splited 4 = 2+2 ; 7 = 4+3 ; 11 = 5+7, is since the difference between the consecutive terms are in AP, if I subtract that difference from the particular term of the sequence I will get the previous term.
This way I can split the whole sequence into an AP sequence and the given sequence with n-1 terms.

this way I can get \(S_n\) and \(S_{n-1}\)
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Re: Help:Tough problem on exponential sequence  [#permalink]

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New post 24 Jul 2010, 21:18
3
I've learnt this method at school, and it can be applied easily to a wide variety of problems tht involve summations.

On observation we find that the 'n'th term of this sequence is of the form

\(T_n=T_{n-1}+ [n-1]\)

for n >= 2 i.e from the 2nd term onwards.

so to find the 60th term (\(T_60\)) all we need to do is this :

\(T_{60} = T_{59} + (60-1) = T_{59}+59\)
\(T_{59} = T_{58}+58\)
and
so
on
.
.
.
.
\(T_3 = T_2+2\)
\(T_2 = T_1+1\)
___________

On summing all the equations' left hand sides and right hand sides, observe that every equations' LHS (starting with the second) cancels with the term in the RHS above it - for example T59 in the second eqn in the LHS cancels with T59 in the RHS of the first equation and so on for the entire system of equations.

After this huge crash of dominoes, all we are left with is

\(T_{60} = T_1 + (1+2+3+4+...59)\)

\(1+2+...59\) as we know is \({59*(59+1)}/2 = 1770\)
and \(T_1 = 1\)

so the answer is 1771 (D)

Just my $0.02.
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Re: Help:Tough problem on exponential sequence  [#permalink]

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New post 25 Jul 2010, 02:56
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Another similar approach

\(S_n = 1+2+4+7+11 ..........T_n\)
\(S_n = 0+1+2+4+ 7 ...........T_{n-1} +T_n\)
Subtract

\(0 = 1+ [(2-1) + (4-2) .......T_n -T_{n-1} ] - T_n\)
\(T_n = 1 + (1+2+3+4.....n-1\))
\(T_n = 1+ n*\frac{(n-1)}{2}\)

=> \(T_{60} = 1+ 59*\frac{60}{2} = 1771\)
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Re: Help:Tough problem on exponential sequence  [#permalink]

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New post 28 Jan 2011, 08:40
gurpreetsingh wrote:
\(T_n = 1+ n*\frac{(n-1)}{2}\)

=> \(T_{60} = 1+ 59*\frac{60}{2} = 1771\)


I understand how you came to this part but I don't understand how you arrived at this part of the solution. Please help me understand.

Thanks in advance!

Mari
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Re: Help:Tough problem on exponential sequence  [#permalink]

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New post 28 Jan 2011, 09:23
I m sry I do not understand what are you asking.
Kindly re-phrase the question:
what you got and what not.
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Re: Help:Tough problem on exponential sequence  [#permalink]

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New post 28 Jan 2011, 11:47
I might missing something small... how u done 1+2+3+4+5+6+7+8+9+10....+59+60
in a short way? can someone please explain the logic?

is there a difference if the amount of numbers is an even or uneven number?

thanks a lot.
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Re: Help:Tough problem on exponential sequence  [#permalink]

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New post 28 Jan 2011, 12:04
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144144 wrote:
I might missing something small... how u done 1+2+3+4+5+6+7+8+9+10....+59+60
in a short way? can someone please explain the logic?

is there a difference if the amount of numbers is an even or uneven number?

thanks a lot.


I m not sure I got your question.

but if you mean how to find value of 1+2+3+4+5.....60

then you may use sum of first n natural numbers = n*(n+1)/2

or you may consider it to be an AP series with a = 1 and d = 1
=> sum = (n/2)*(2a+(n-1)*d)
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Re: Help:Tough problem on exponential sequence  [#permalink]

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New post 28 Jan 2011, 12:16
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144144 wrote:
I might missing something small... how u done 1+2+3+4+5+6+7+8+9+10....+59+60
in a short way? can someone please explain the logic?

is there a difference if the amount of numbers is an even or uneven number?

thanks a lot.


First of all we have 1+(1+2+3+...+59).

Next, the sum of the elements in any evenly spaced set is given by: \(Sum=\frac{first+last}{2}*# \ of \ terms\), the mean multiplied by the number of terms. So, \(1+(1+2+3+...+59)=1+\frac{1+59}{2}*59=1+1770=1771\).

Check this thread for more: math-number-theory-88376.html also this thread: sequences-progressions-101891.html
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Re: Help:Tough problem on exponential sequence  [#permalink]

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New post 28 Jan 2011, 12:41
Bunuel wrote:
144144 wrote:
I might missing something small... how u done 1+2+3+4+5+6+7+8+9+10....+59+60
in a short way? can someone please explain the logic?

is there a difference if the amount of numbers is an even or uneven number?

thanks a lot.


First of all we have 1+(1+2+3+...+59).

Next, the sum of the elements in any evenly spaced set is given by: \(Sum=\frac{first+last}{2}*# \ of \ terms\), the mean multiplied by the number of terms. So, \(1+(1+2+3+...+59)=1+\frac{1+59}{2}*59=1+1770=1771\).

Check this thread for more: math-number-theory-88376.html also this thread: sequences-progressions-101891.html

Now i get it. Thanks again Bunuel! Very helpful :)
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Re: What is the sixtieth term in the following sequence? 1, 2, 4  [#permalink]

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Re: What is the sixtieth term in the following sequence? 1, 2, 4   [#permalink] 23 Nov 2018, 04:46
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