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n is an integer greater than or equal to 0. The sequence tn

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n is an integer greater than or equal to 0. The sequence tn [#permalink] New post   Updated on: 19 Jul 2021, 13:45
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n is an integer greater than or equal to 0. The sequence \(t_n\) for n > 0 is defined as \(t_n = t_{n-1} + n\). Given that \(t_0 = 3\), is tn even?

(1) n + 1 is divisible by 3
(2) n - 1 is divisible by 4
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B

Originally posted by Onell on 15 Mar 2011, 22:55.
Last edited by Bunuel on 19 Jul 2021, 13:45, edited 3 times in total.
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Re: n is an integer greater than or equal to 0. The sequence tn [#permalink] New post   16 Mar 2011, 02:00
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See, t0 is odd.
Then we add an odd number (n=1), so t1 is even.
Then we add an even number(n=2), so t2 is even.
Then we add an odd number (n=3), so t3 is odd.
Then we add an even number(n=4), so t4 is odd.
Then we add an odd number(n=5), so t5 is even.
Then we add an even number(n=6), so t6 is even.
Then we add an odd number(n=7), so t7 is odd.
Then we add an even number(n=8), so t8 is odd.
Etc........

As you could see here the sequence is connected with divisibility by 4.
So (1) tells us about divisibility by 3 and it should be not sufficient. See countrexample: t2 is even (n+1 is 3), t8 is odd (n+1=9)
(2) alone should be sufficient since it tells us about the divisibility by 4, and we see that n which are divided by 4 with remainder of 1 or 2 is even. If the remainder is 0 or 3, it is odd. Here the reminder is 1 (n-1 is divided by 4), so the term should be even.

The answer is (B)
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Re: n is an integer greater than or equal to 0. The sequence tn [#permalink] New post   16 Mar 2011, 04:57
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t0 = 3,

So t1 = t0 + n = 3 + n

t2 = 3 + n + n = 3 + 2n

t3 = 3 + 2n + n = 3 + 3n

So tn is even only if n is odd as odd + odd, if n is even then it will be odd + even = odd


From (1) we have n + 1 is divisible by 3 => n + 1 = 3k where k is an integer.

So n = 3k - 1 which can be values like 2, 5, 8, so (1) is insufficient.

From (2) we have n-1 = 4m where m is an integer

So n = 4m + 1, which is always odd

Hence (2) is sufficient, so the answer is B.
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Re: n is an integer greater than or equal to 0. The sequence tn [#permalink] New post   24 Nov 2013, 15:22
It should be E...n-1 being divisible by 4, based on the chart, that means it's either 4 (as you can see if it's 4, then the next term in the sequence is 6), or 24 for (in which case the next term in the series is 31). OA is incorrect.
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Re: n is an integer greater than or equal to 0. The sequence tn [#permalink] New post   24 Nov 2013, 15:40
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AccipiterQ wrote:
n is an integer greater than or equal to 0. The sequence \(t_n\) for n > 0 is defined as \(t_n = t_{n-1} + n\). Given that \(t_0 = 3\), is tn even?

(1) n + 1 is divisible by 3
(2) n - 1 is divisible by 4

It should be E...n-1 being divisible by 4, based on the chart, that means it's either 4 (as you can see if it's 4, then the next term in the sequence is 6), or 24 for (in which case the next term in the series is 31). OA is incorrect.


The OA is correct.

(2) n - 1 is divisible by 4 means that \(n=4k+1\), thus n is 1, 5, 9, ...

\(t_1=4=even\).
\(t_5=18=even\).
\(t_9=48=even\).
...

All are even.
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Re: n is an integer greater than or equal to 0. The sequence tn [#permalink] New post   28 Dec 2013, 09:34
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First off let's see the sequence behavior charting some numbers.

t(0)=3+0 = O
t(1)=3+1 = E
t(2)=4+2 = E
t(3)=6+3 = O
t(4)=9+4 = O
t(5)=13+5= E
t(8) = O

we can notice a repeating pattern (E, E, O, O) we need to figure out how n relates to a multiple of 4.

st1 n could be 2, 5, 8, 11, 14 etc.. checking the chart we can tell that this statement is not sufficient
st2 tells us how n relates to a multiple of 4 and indeed if we plug some values in we can safely claim that t(n) is even.
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Re: n is an integer greater than or equal to 0. The sequence tn [#permalink] New post   05 Jan 2014, 08:26
I have a doubt here, 2nd part says that it is divisible by 4 then why have you wrote the expression as n= 4k+1. it doesn't talks about remainder here right? isn't n=4k enough?

Bunuel wrote:
AccipiterQ wrote:
n is an integer greater than or equal to 0. The sequence \(t_n\) for n > 0 is defined as \(t_n = t_{n-1} + n\). Given that \(t_0 = 3\), is tn even?

(1) n + 1 is divisible by 3
(2) n - 1 is divisible by 4

It should be E...n-1 being divisible by 4, based on the chart, that means it's either 4 (as you can see if it's 4, then the next term in the sequence is 6), or 24 for (in which case the next term in the series is 31). OA is incorrect.


The OA is correct.

(2) n - 1 is divisible by 4 means that \(n=4k+1\), thus n is 1, 5, 9, ...

\(t_1=4=even\).
\(t_5=18=even\).
\(t_9=48=even\).
...

All are even.
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Re: n is an integer greater than or equal to 0. The sequence tn [#permalink] New post   05 Jan 2014, 09:57
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rgyanani wrote:
I have a doubt here, 2nd part says that it is divisible by 4 then why have you wrote the expression as n= 4k+1. it doesn't talks about remainder here right? isn't n=4k enough?

Bunuel wrote:
AccipiterQ wrote:
n is an integer greater than or equal to 0. The sequence \(t_n\) for n > 0 is defined as \(t_n = t_{n-1} + n\). Given that \(t_0 = 3\), is tn even?

(1) n + 1 is divisible by 3
(2) n - 1 is divisible by 4

It should be E...n-1 being divisible by 4, based on the chart, that means it's either 4 (as you can see if it's 4, then the next term in the sequence is 6), or 24 for (in which case the next term in the series is 31). OA is incorrect.


The OA is correct.

(2) n - 1 is divisible by 4 means that \(n=4k+1\), thus n is 1, 5, 9, ...

\(t_1=4=even\).
\(t_5=18=even\).
\(t_9=48=even\).
...

All are even.


n - 1 is divisible by 4 --> \(n-1=4k\) --> \(n=4k+1\) --> n is 1 more than a multiple of 4.
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Re: n is an integer greater than or equal to 0. The sequence tn [#permalink] New post   07 Jan 2014, 02:14
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Hi
Bunuel pl review the logic below

its a AP where d = tn-tn-1= n where 3 is the first term

Now the tn = 3+ (n-1)n {by formula tn= a+(n-1)d}

in this equation n and n-1 are consecutive numbers

for statement #1: n+1, which is next consecutive number in the sequence, is divisible by 3, but we don't know whether its even or odd ( including 3,6,9,..) so insuff

for statement #2: n-1 is divisible by 4 so n-1 is even hence n is odd and n(n-1) is even. And 3 + even = odd suff

thanks
sid
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Re: n is an integer greater than or equal to 0. The sequence tn [#permalink] New post   18 Sep 2016, 08:51
Hi Bunnel,

If we apply the nth term of an AP formula,
we have
tn= 3 + (n-1)* d
In this AP, (this is an AP because tn -tn-1=n)
we get, tn=3 + (n-1)*n
now because (n-1)*n is always even (product of two consecutive numbers is always even), we get
tn= 3 + even = odd.
So clearly, tn is odd so (A) should be sufficient.

Moving to B, (n-1) is divisible by 4, that means n-1 is even so n is odd,
again tn = 3+n-1*n = 3+ even = odd

Why is th e answer not D?
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Re: n is an integer greater than or equal to 0. The sequence tn [#permalink] New post   18 Sep 2016, 09:45
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gauravprashar17 wrote:
Hi Bunnel,

If we apply the nth term of an AP formula,
we have
tn= 3 + (n-1)* d
In this AP, (this is an AP because tn -tn-1=n)
we get, tn=3 + (n-1)*n
now because (n-1)*n is always even (product of two consecutive numbers is always even), we get
tn= 3 + even = odd.
So clearly, tn is odd so (A) should be sufficient.

Moving to B, (n-1) is divisible by 4, that means n-1 is even so n is odd,
again tn = 3+n-1*n = 3+ even = odd

Why is th e answer not D?


The given sequence is NOT an arithmetic progression. You'd notice it if you'd try to write down several terms:
\(t_0=3\)
\(t_1=t_0+1=4\)
\(t_2=t_1+2=6\)
\(t_3=t_2+3=9\)
\(t_4=t_3+4=13\)
\(t_5=t_4+5=18\)
\(t_6=t_5+6=24\)
\(t_7=t_6+7=31\)
\(t_8=t_7+8=39\)

...

(1) n + 1 is divisible by 3 --> n + 1 = 3x --> n = 3x - 1, thus n can be 2, 5, 8, ... tn ca be even (for example, t2) and odd (for example, t8).
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Re: n is an integer greater than or equal to 0. The sequence tn [#permalink] New post   08 Jan 2017, 00:19
Bunuel please help:

I get why (1) is NS:
- T_0 => yes N even
- T_1 => No N not even

I'm confused on (2):

- if T_7 then no N = odd
- if T_2 then yes N = odd

where is my logic off? Thanks!!

Bunuel wrote:
gauravprashar17 wrote:
Hi Bunnel,

If we apply the nth term of an AP formula,
we have
tn= 3 + (n-1)* d
In this AP, (this is an AP because tn -tn-1=n)
we get, tn=3 + (n-1)*n
now because (n-1)*n is always even (product of two consecutive numbers is always even), we get
tn= 3 + even = odd.
So clearly, tn is odd so (A) should be sufficient.

Moving to B, (n-1) is divisible by 4, that means n-1 is even so n is odd,
again tn = 3+n-1*n = 3+ even = odd

Why is th e answer not D?


The given sequence is NOT an arithmetic progression. You'd notice it if you'd try to write down several terms:
\(t_0=3\)
\(t_1=t_0+1=4\)
\(t_2=t_1+2=6\)
\(t_3=t_2+3=9\)
\(t_4=t_3+4=13\)
\(t_5=t_4+5=18\)
\(t_6=t_5+6=24\)
\(t_7=t_6+7=31\)
\(t_8=t_7+8=39\)

...

(1) n + 1 is divisible by 3 --> n + 1 = 3x --> n = 3x - 1, thus n can be 2, 5, 8, ... tn ca be even (for example, t2) and odd (for example, t8).
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Re: n is an integer greater than or equal to 0. The sequence tn [#permalink] New post   08 Jan 2017, 05:28
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mdacosta wrote:
Bunuel please help:

I get why (1) is NS:
- T_0 => yes N even
- T_1 => No N not even

I'm confused on (2):

- if T_7 then no N = odd
- if T_2 then yes N = odd

where is my logic off? Thanks!!

Bunuel wrote:
gauravprashar17 wrote:
Hi Bunnel,

If we apply the nth term of an AP formula,
we have
tn= 3 + (n-1)* d
In this AP, (this is an AP because tn -tn-1=n)
we get, tn=3 + (n-1)*n
now because (n-1)*n is always even (product of two consecutive numbers is always even), we get
tn= 3 + even = odd.
So clearly, tn is odd so (A) should be sufficient.

Moving to B, (n-1) is divisible by 4, that means n-1 is even so n is odd,
again tn = 3+n-1*n = 3+ even = odd

Why is th e answer not D?


The given sequence is NOT an arithmetic progression. You'd notice it if you'd try to write down several terms:
\(t_0=3\)
\(t_1=t_0+1=4\)
\(t_2=t_1+2=6\)
\(t_3=t_2+3=9\)
\(t_4=t_3+4=13\)
\(t_5=t_4+5=18\)
\(t_6=t_5+6=24\)
\(t_7=t_6+7=31\)
\(t_8=t_7+8=39\)

...

(1) n + 1 is divisible by 3 --> n + 1 = 3x --> n = 3x - 1, thus n can be 2, 5, 8, ... tn ca be even (for example, t2) and odd (for example, t8).


(2) n - 1 is divisible by 4 means that \(n=4k+1\), thus n is 1, 5, 9, ... So, n cannot be 2 or 7.

\(t_1=4=even\).
\(t_5=18=even\).
\(t_9=48=even\).
...

All are even.
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Re: n is an integer greater than or equal to 0. The sequence tn [#permalink] New post   23 Feb 2017, 05:46
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PROMPT ANALYSIS
n is a natural number and tn follows the sequence as tn = tn-1 + n with t0 = 3.
Therefore, t1 =4,t2 =6, t3=9,t4 =13 t5 = 18 t6 = 24 t7 = 31 t8 = 39 t9= 48 t10 =58 t11 = 69…..

Superset
The answer will be either YES or NO.

Translation
In order to know if tn is even we need:
1# exact value of n
2# any equation that will conclude the question

Statement analysis

St 1: n +1 = 3k; n = 3k-1. So we are talking about t2(even), t5(even), t8(odd), t11(odd). NO CONCLUSION. Hence option a, d eliminated.

St 2: n - 1 = 4j; n = 4j +1. So we are talking about t1(even), t5(even), t9(even) t13(even). Hence it is even. Answr.

Option B
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Re: n is an integer greater than or equal to 0. The sequence tn [#permalink] New post   19 Jul 2022, 19:06
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Re: n is an integer greater than or equal to 0. The sequence tn [#permalink]
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