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

Question

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

Bunuel · Accepted Answer

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.

prdatta · Answer

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}+

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.

gurpreetsingh · Answer

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

gurpreetsingh · Answer

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}

gurpreetsingh · Answer

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+   - 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

mariyea · Answer

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

gurpreetsingh · Answer

I m sry I do not understand what are you asking.
Kindly re-phrase the question:
what you got and what not.

144144 · Answer

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.

gurpreetsingh · Answer

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)

Bunuel · Answer

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

mariyea · Answer

Now i get it. Thanks again Bunuel! Very helpful

gracie · Answer

note sequence is variant of handshake sequence
handshake sequence: t↓n=n(n-1)/2
variant: t↓n= +1
t↓60= +1=1771
D

GMATinsight · Answer

Please find the video solution and solution in the screenshot as attached

https://www.youtube.com/watch?v=c7b4POllgtM

jaykayes · Answer

Here's how to find the sum of "n" consecutive terms of any finite arithmetic progression.

We'll use the following example:
1, 3, 5, ....... 21, 23, 25.

The sum of the first and last terms is 26.
The sum of the second and second-last terms is also 26.
The sum of the third and third-last terms is also 26.
Similarly, as you go inwards, every pair of terms will add up to 26.

So the sum = (no. of pairs) x 26
= (half the number of terms) x 26
= (half the number of terms) x (sum of the first and last terms)
= (n/2) x (F + L)

Incidentally, this will work whether the number of terms you're adding is even or odd!

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

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