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# If sequence T is defined for all positive integers n such that tn +1 =

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Manager
Joined: 14 Oct 2014
Posts: 66
Location: United States
GMAT 1: 500 Q36 V23
If sequence T is defined for all positive integers n such that tn +1 =  [#permalink]

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Updated on: 13 Jan 2015, 15:31
3
8
00:00

Difficulty:

55% (hard)

Question Stats:

70% (03:15) correct 30% (03:19) wrong based on 209 sessions

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If sequence T is defined for all positive integers n such that $$t_{(n +1)} = t_{n} + n$$, and $$t_3 = 14$$, what is $$t_{20}$$?

A. 101
B. 187
C. 191
D. 201
E. 251

Originally posted by viktorija on 12 Jan 2015, 14:18.
Last edited by viktorija on 13 Jan 2015, 15:31, edited 1 time in total.
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Re: If sequence T is defined for all positive integers n such that tn +1 =  [#permalink]

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12 Jan 2015, 21:12
4
3
viktorija wrote:
If sequence T is defined for all positive integers n such that tn +1 = tn + n, and t 3 = 14, what is t 20?
A. 101
B. 187
C. 191
D. 201
E. 251

Please ensure that the question is typed clearly and has no ambiguity. You are given t(n +1) = t(n) + n not t(n +1) = t(n + n)

$$t_{n+1} = t_{n} + n$$
$$t_{20} = t_{19} + 19$$
$$t_{20} = t_{18} + 18 + 19$$

We see the pattern:
$$t_{20} = t_3 + 3 + 4 + 5 + 6 + ... + 17 + 18 + 19$$

$$t_{20} = 14 + 3 + 4 + 5 + ... + 18 + 19$$
Split 14 into 11 + 1 + 2 for ease of calculation:

$$t_{20} = 11 + 1 + 2 + 3 + 4 + ... + 18 + 19$$
Sum of first m positive integers is m(m+1)/2 so

$$t_{20} = 11 + 19*20/2 = 201$$

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Posts: 8
Re: If sequence T is defined for all positive integers n such that tn +1 =  [#permalink]

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12 Jan 2015, 15:38
2
3
ok, not sure if this is the best explanation, but here is my way of solving this :

tn +1 = tn + n,
t 3 = 14
so:
t 4 = t 3 + 3
t 5 = t 4 + 4, so we replace t 4 with the previous result : t 5 = (t 3 + 3 ) + 4
....
so we get
t 20 = t 3 + 3 + 4 + 5 + 6....+ 19
the 3 + 4 + 5 + ... + 19 sequence is equal the sequence 1 + 2 + ... + 19, minus 1 + 2 at the beginning (so, minus 3)
and the 1 + 2 ... + 19 sequence is equal to n*(n+1)/2 with n being 19, so equal to 19 * 20 / 2 = 190
so :
t 20 = t 3 + 190 - 3
t 20 = 14 + 190 - 3 = 190 + 11 = 201, hence answer D
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Joined: 30 Oct 2011
Posts: 190
Re: If sequence T is defined for all positive integers n such that tn +1 =  [#permalink]

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15 Jan 2015, 03:57
1
1
viktorija wrote:
If sequence T is defined for all positive integers n such that t(n +1) = t(n) + n, and t 3 = 14, what is t 20?
A. 101
B. 187
C. 191
D. 201
E. 251

Calculate the first few values to see a pattern

t1= 11
t2= 12
t3= 14
t4= 17
t5= 21

The sequence is basically like this t1 + 1 +2 +3 +4 +5 ...

So t20 = t1 + 1 +2 +3 up to 19 = 11 + [(19 +1)/2] * 19 = 201

Hope this helps. A big thanks to Bunuel, I learned this technique reading his posts.
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Joined: 14 Nov 2016
Posts: 1316
Location: Malaysia
Re: If sequence T is defined for all positive integers n such that tn +1 =  [#permalink]

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23 Mar 2017, 17:20
1
2
viktorija wrote:
If sequence T is defined for all positive integers n such that $$t_{n +1} = t_{n} + n$$, and $$t_3 = 14$$, what is $$t_{20}$$ ?

A. 101
B. 187
C. 191
D. 201
E. 251

Source : Manhattan Advanced Quant Question No. 8

OFFICIAL SOLUTION

First, we can write a few of the terms, noting the relationship of each term to the previous one:

$$t_3 = 14$$
$$t_4 = 14 + 3$$
$$t_5 = 14 + 3 + 4$$
$$t_6 = 14 + 3 + 4 + 5$$

$$t_{20} = 14 + (3 + 4 + 5 + … + 18 + 19)$$

To evaluate $$t_{20}$$ , we need to compute the sum contained in parentheses above.

We can use the rule that (sum of a set of consecutive integers) = (middle term) × (number of terms).
The middle term can be found by taking the average of the two extreme terms 3 and 19 to get 11. The number of terms is $$19 – 3 + 1 = 17$$.
Now we can compute $$11 × 17 = 187$$.

Finally, we have $$t_{20} = 14 + (3 + 4 + 5 + … + 18 + 19) = 14 + 187 = 201$$.

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Joined: 13 Dec 2013
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Re: If sequence T is defined for all positive integers n such that tn +1 =  [#permalink]

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12 Apr 2017, 16:49
BrainFog wrote:
viktorija wrote:
If sequence T is defined for all positive integers n such that t(n +1) = t(n) + n, and t 3 = 14, what is t 20?
A. 101
B. 187
C. 191
D. 201
E. 251

Calculate the first few values to see a pattern

t1= 11
t2= 12
t3= 14
t4= 17
t5= 21

The sequence is basically like this t1 + 1 +2 +3 +4 +5 ...

So t20 = t1 + 1 +2 +3 up to 19 = 11 + [(19 +1)/2] * 19 = 201

Hope this helps. A big thanks to Bunuel, I learned this technique reading his posts.

Where does the "11" come from here?
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Joined: 06 Feb 2016
Posts: 8
Re: If sequence T is defined for all positive integers n such that tn +1 =  [#permalink]

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19 Apr 2017, 22:35
$$t_{n+1} = t_{n} + n$$
$$t_{20} = t_{19} + 19$$
$$t_{20} = t_{18} + 18 + 19$$

Also, since $$t_{3} = t_{2} + 2 = 14$$ ->
$$t_{2} = t_{1} + 1$$
and $$t_{1} =11$$

The pattern:
$$t_{20} = t_1 + (1 +2 + 3 + 4 + 5 + 6 + ... + 17 + 18 + 19)$$ where t_1 = 11

Sum of numbers from 1 to 19 inclusive is Avg. * # of terms
Avg. = (1 + 19)/2 = 10
Sum = 10*19=190

Thus:
$$t_{20} = t_1 + 190 = 11+190 = 201$$

Senior Manager
Joined: 06 Dec 2016
Posts: 250
If sequence T is defined for all positive integers n such that tn +1 =  [#permalink]

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20 Apr 2017, 10:22
My approach:
t3 = 14
t4 = 14 + 3 = 17
t20 = 14+3 + 4+5+ ... + 19
t20= 11 + 1+ 2 + ... + 19

sum of (1 to 19) = 19/2(19 + 1) = 190

Therefore

t20 = 190 + 11 = 201 Which is answer D
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Posts: 8776
Re: If sequence T is defined for all positive integers n such that tn +1 =  [#permalink]

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24 Jul 2018, 05:04
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Re: If sequence T is defined for all positive integers n such that tn +1 = &nbs [#permalink] 24 Jul 2018, 05:04
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