see above sir

Carcass yes, I saw but the sequence repeats after 5 terms not 6 ... Also, can we not use Number of terms x Median formula here? Thanks!

No Sir

the pattern above is starting all over again at 6th term which is zero The above explanation is correct.

No sir

We do not have information about the median. The question is conceived to test a recursive sequence

Let’s first list some terms to see if we can spot a pattern…
\(t_ 1 = 1\)
\(t_ 2 = 1\)
\(t_ 3 = t_2 – t_1 = 1 – 1 = 0\)
\(t_ 4 = t_3 – t_2 = 0 – 1 = -1\)
\(t_ 5 = t_4 – t_3 = (-1) – 0 = -1\)
\(t_ 6 = t_5 – t_4 = (-1) – (-1) = 0\)
\(t_ 7 = t_6 – t_5 = 0 - (-1) = 1\)
\(t_ 8 = t_7 – t_6 = 1 - 0 = 1\)
\(t_ 9 = t_8 – t_7 = 1 - 1 = 0\)
\(t_ {10} = t_9 – t_8 = 0 - 1 = -1\)
.
.
.
At this point, we can see that the pattern repeats itself after 6 terms
In other words, the six terms from \(t_1\) to \(t_6\) are exactly the same as the six terms from \(t_7\) to \(t_{12} \) (as well as the six terms from \(t_{13}\) to \(t_{18}\), etc)

The sum of the first six terms \(= 1 + 1 + 0 + (-1) + (-1) + 0 = 0\)
So, the sum of the next six terms after that \(= 0\)
And the sum of the next six terms after that \(= 0\)
Etc.

There are 8 sets of six terms from \(t_1\) to \(t_{48}\), which means the sum of the first 48 terms \(= (8)(0) = 0\)
All we need to do now is add \(t_ {49} \) and \(t_ {50} \)

We can follow the pattern to see that \(t_ {49} = 1\) and \(t_ {50} = 1\), which means the sum of the first 50 terms = 2

Answer: E

Since the question says n>2, shouldn't we be checking from t3 to t52?

The batch of certain terms gets repeated in the series hence we don't need to calculate till \(t_{52}\).

t1 and t2 are still part of the sequence.
So they will be included in the sum of terms from t1 to t50

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