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Since Tn = Tn-1 + n and T0 = 3
T1 = 3 + 1 = 4, T2 = 4 + 2 = 6,
T3 = 6 + 3 = 9, T4 = 9 + 4 =13,
T5 = 13 + 5 = 18, T6 = 18 + 6 = 24,
T7 = 24 + 7 = 31, T8 = 31+8 = 39,
T9 = 39 + 9 = 48, T10 = 48 + 10 = 58,
T11 = 58 + 11 = 69, T12 = 69 + 12 = 81,
T13 = 81 + 13 = 94

The pattern formed is as follows : E,E,O,O,E,E,O,O,E,E,O,O,E.....

1. If n+1 is divisble by 3, n could be 2,5,8.. and so on.
From the question stem T2 and T5 are even but, T8 is odd
Hence, insufficient
2. n-1 is divisible by 4
If n-1 is divisible by 4, n could be 5,9,13... and so on.
From the question stem, T5,T9,T13 are all even. Sufficient(Option B)
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amsey1382
Q. n is an integer greater than or equal to 0. The sequence Tn for n>0 is defined by Tn = Tn-1 + n. If T0 = 3, is nth Term, i.e. Tn even ?

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

This question is discussed here: n-is-an-integer-greater-than-or-equal-to-0-the-sequence-tn-110969.html

Hope it helps.



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