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n is an integer greater than or equal to 0. The sequence tn [#permalink]
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15 Mar 2011, 22:55
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n is an integer greater than or equal to 0. The sequence \(t_n\) for n > 0 is defined as \(t_n = t_{n1} + n\). 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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Last edited by Bunuel on 24 Nov 2013, 15:33, edited 2 times in total.
Added the OA.



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Re: Sequence [#permalink]
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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 (n1 is divided by 4), so the term should be even. The answer is (B)
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Re: Sequence [#permalink]
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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 n1 = 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]
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24 Nov 2013, 15:22
It should be E...n1 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]
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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 \(t_n = t_{n1} + n\). 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...n1 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]
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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]
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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 \(t_n = t_{n1} + n\). 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...n1 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]
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05 Jan 2014, 09:57
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 \(t_n = t_{n1} + n\). 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...n1 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 > \(n1=4k\) > \(n=4k+1\) > 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]
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07 Jan 2014, 02:14
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Hi Bunuel pl review the logic below
its a AP where d = tntn1= n where 3 is the first term
Now the tn = 3+ (n1)n {by formula tn= a+(n1)d}
in this equation n and n1 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: n1 is divisible by 4 so n1 is even hence n is odd and n(n1) 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]
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25 May 2014, 22:38
n is an integer greater than or equal to 0. The sequence t_n for n > 0 is defined as t_n = t_{n1} + n. Given that t_0 = 3, is tn even?
(1) n + 1 is divisible by 3 (2) n  1 is divisible by 4 T_0=3 T_1=3+1=4 T_2=4+2=3+1+2=6 T_3=6+3=3+1+2+3=9 T_4=9+4=3+1+2+3+4
T_n= 3+ (sum of integers from 1 to n)
T_n can only be even if the sum of integers from 1 to n is odd, as odd (3)+ odd=even
1. n+1 can only be divisible by 3 if T_n= T_2, T_5, T_8, T_11.... sum of integers from 1 to 2=3 (odd) sum of integers from 1 to 5=15 (odd) sum of integers from 1 to 8=36 (even)
Not sufficient
2. n1 can only be divisible by 4 if T_n= T_5, T_9, T_13, T_17..... sum of integers from 1 to 5=15 (odd) sum of integers from 1 to 9=45 (odd) sum of integers from 1 to 13= 91 (odd) sum of integers from 1 to 17=153 (odd)
Sufficient
Ans. B



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Re: n is an integer greater than or equal to 0. The sequence tn [#permalink]
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07 Jul 2014, 08:55
gmat6nplus1 wrote: 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. how exactly does st.1 tell us n could be 2, 5, 8 ,11 ,14 could you show the plugging in?



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Re: n is an integer greater than or equal to 0. The sequence tn [#permalink]
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25 Jan 2015, 05:22
sagnik2422 wrote: gmat6nplus1 wrote: 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. how exactly does st.1 tell us n could be 2, 5, 8 ,11 ,14 could you show the plugging in? Statement says 3k=n+1 so n =3k1, putting k=1 we get n=2, k=2 gives n=5 , k=3 gives n=8 and so on......
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Re: n is an integer greater than or equal to 0. The sequence tn [#permalink]
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29 May 2016, 07:38
Hi,
Any help would be appreciated. I just can't seem to get my head around this question.
1. I don't understand what is the meaning of "is Tn even"? How do we even prove that. Further what help are both the statements providing in helping to prove it.
Below mentioned is all i understand and gather from the question. Upon substitution of the values, we get Tn as follows 
t(0)=3+0 = 3 = ODD t(1)=3+1 = 4 = EVEN t(2)=4+2 = 6 = EVEN t(3)=6+3 = 9 = ODD t(4)=9+4 = 13 = ODD t(5)=13+5= 18 = EVEN t(6) =18 + 6 = 24 = EVEN
Have no clue what to do after this.
Thanks in advance



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Re: n is an integer greater than or equal to 0. The sequence tn [#permalink]
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18 Sep 2016, 08:51
Hi Bunnel,
If we apply the nth term of an AP formula, we have tn= 3 + (n1)* d In this AP, (this is an AP because tn tn1=n) we get, tn=3 + (n1)*n now because (n1)*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, (n1) is divisible by 4, that means n1 is even so n is odd, again tn = 3+n1*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]
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18 Sep 2016, 09:45
gauravprashar17 wrote: Hi Bunnel,
If we apply the nth term of an AP formula, we have tn= 3 + (n1)* d In this AP, (this is an AP because tn tn1=n) we get, tn=3 + (n1)*n now because (n1)*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, (n1) is divisible by 4, that means n1 is even so n is odd, again tn = 3+n1*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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n is an integer greater than or equal to 0. The sequence tn [#permalink]
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24 Sep 2016, 02:23
Hi Buneul, Kindly correct me if I am right.
Given t0=3 then t1=3+1 t2= 3+1+2 t3=3+1+2+3 t4=3+1+2+3+4 and so on So the general term for tn= 3+sum of n natural numbers => tn=3 + n(n+1)/2........(1)As per this only if either n or n+1 is a multiple of 4, will the term n(n+1)/2 be even, otherwise not.
Now, as per the statements 1) n+1 is divisible by 3 => tn is divisible by 3 but we cant determine whether the 2nd term of eqn (1) is even or odd. INSUFFICIENT 2) n1 is divisible by 4 => n1 is even, n is odd and n1+2=n+1 is also even but not a multiple of 4 => (n+1)/2 is odd So, the 2nd term of eqn (1) is odd. Hence, tn=odd+ (oddxodd)= even. SUFFICIENT
Answer is B



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Re: n is an integer greater than or equal to 0. The sequence tn [#permalink]
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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 + (n1)* d In this AP, (this is an AP because tn tn1=n) we get, tn=3 + (n1)*n now because (n1)*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, (n1) is divisible by 4, that means n1 is even so n is odd, again tn = 3+n1*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]
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08 Jan 2017, 05:28
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 + (n1)* d In this AP, (this is an AP because tn tn1=n) we get, tn=3 + (n1)*n now because (n1)*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, (n1) is divisible by 4, that means n1 is even so n is odd, again tn = 3+n1*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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n is an integer greater than or equal to 0. The sequence tn [#permalink]
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08 Jan 2017, 05:54
After evaluating for different values of n using equation \(t_n = t_{n1} + n\) we get sequence: 3,4,5,6,7,8......
which shows that , when n is odd, the corresponding term,tn is odd and when n is even ,the corresponding term, tn is even We need to find : n is even or odd
statement(1) : n + 1 is divisible by 3 there are two possibilities of n , odd and even (as multiples of 3 can be odd or even) Ex. n=5 , n+1=6 : divisible by 3 n=2 , n+1=3 : divisible by 3 n could be odd or even >> not sufficient.
statement(2) : n  1 is divisible by 4 for n1 to be divisible by 4, n1 has to be even (as multiples of 4 are always even) Hence, n is odd. >> sufficient.
Therefore , tn is odd.
Ans : B




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