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willget800
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What is the sixtieth term in the following sequence? 1, 2, 25 Apr 2006, 19:04
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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

Note: This is from the MGMAT Web site.

How to even approach this question?



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What is the sixtieth term in the following sequence? 1, 2, 25 Apr 2006, 19:24
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T1 = 1
T2 = 2 = T1+1
T3 = 4 = T1+3 = T1+1+2
T4 = 7 = T1+6 = T1+1+2+3
T5 = 11 = T1+10 = T1+1+2+3+4
T6 = 16 = T1+15 = T1+1+2+3+4+5
T7 = 22 = T1+21 = T1+1+2+3+4+5+6
T8 = 30 = T1+29 = T1+1+2+3+4+5+6+7

So T60 = T1+1+2+3....+59

Sum of digits from 1 to 59 = n/2(a+l) = 59/2(1+59) = 1770
T60 = 1771+1 = 1771

Ans should be D
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Charlie45
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What is the sixtieth term in the following sequence? 1, 2, 25 Apr 2006, 19:36
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Quote:
Sum of digits from 1 to 59 = n/2(a+l) = 59/2(1+59) = 1770
T60 = 1771+1 = 1771
Show more


Could you briefly elaborate on the sum of digits calculation? Thx
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willget800
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What is the sixtieth term in the following sequence? 1, 2, 25 Apr 2006, 19:38
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tht was too good to solve it.... I spend 20 mins.. gave it up .. if you did this under 2 minds.. uare indeed a genious! too good of a solution..

how to appraoch these problems?
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What is the sixtieth term in the following sequence? 1, 2, 26 Apr 2006, 02:16
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Charlie45
Quote:
Sum of digits from 1 to 59 = n/2(a+l) = 59/2(1+59) = 1770
T60 = 1771+1 = 1771

Could you briefly elaborate on the sum of digits calculation? Thx
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What we have is an Arithmetic progression series: 1,2,3,4... to 59. The sum for such a series is defined as:

S = n/2 (a+l) where n = number of terms, a=first term, l=last term.

So for our sum of 1+2+3...+59, we have S = 59/2(1+59)=1770

There's another equation that can be used to solve for the sum of an A.P. That's

S = n/2(2a + [n-1]d) where n = number of terms, a = first term, d is the arithmetic difference (for our case, it's 1)

So S = 59/2 (2(1) + [59-1]1) = 59/2(60) = 1770.

Hope that helps.
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What is the sixtieth term in the following sequence? 1, 2, 26 Apr 2006, 02:18
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willget800
tht was too good to solve it.... I spend 20 mins.. gave it up .. if you did this under 2 minds.. uare indeed a genious! too good of a solution..

how to appraoch these problems?
Show more


Normally, if they're asking for a large term, like 60th term, or 1000th term etc, there's some arrangment invovled and I would look at finding a common difference between running terms, and try to tie them to the first term.
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What is the sixtieth term in the following sequence? 1, 2, 26 Apr 2006, 02:27
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Yeah got D.. used the same approah...

Good question!



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