If $n$ th term of the series $4 + 5 + 8 + 13 + 20 + 29 + . . .$ and $89 + 95 + 101 + 107 + . . .$ are equal, then if you think $n$ exists write $1$ and if you think $n$ doesn't exist then write $0.$

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Solution

If $n$ the term of the series $4 + 5 + 8 + 13 + 20 + 29 + . . .$ and $89 + 95 + 101 + 107 + . . .$ are equal
$S_{1} = 4 + 5 + 8 + 13 + 20 + 29 + . . . .$
$S_{1} = 4 + 5 + 8 + 13 + 20 + . . . .$
Subtract both we get
$a_{n} = 4 + 1 + 3 + 5 + 7 + . . . . . . . .$
$a_{n} - 4 = \frac{(n - 1)}{2} (2 + (n - 1 - 1) 2)$
$a_{n} = 4 + (n - 1)^{2}$
$S_{2} = 89 + 95 + 101 + 107 + . . .$
$S_{2} = 89 + 95 + 101 + 107 + . . .$
Subtract both we get
$b_{n} = 89 + 6 + 6 + 6 + 6 + . . . . . . . .$
$b_{n} - 89 = (n - 1) 6$
$b_{n} = 89 + 6 n - 6$
$b_{n} = 6 n + 83$
$b_{n} = a_{n}$
Then $6 n + 83 = 4 + (n - 1)^{2}$
$n^{2} - 8 n - 78 = 0$
$n = 13.69 a n d n = - 5.69$
So $n$ is not an integer
so the answer is false or $0$

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