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praveengmat
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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gurpreetsingh
praveengmat
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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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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gurpreetsingh
\(T_n = 1+ n*\frac{(n-1)}{2}\)

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

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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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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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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144144
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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144144
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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Bunuel
144144
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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praveengmat
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

note sequence is variant of handshake sequence
handshake sequence: t↓n=n(n-1)/2
variant: t↓n=[n(n-1)/2]+1
t↓60=[(60*59)/2]+1=1771
D
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praveengmat
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

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


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144144
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.


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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