Which term of the sequence \[25,24\dfrac{1}{4},23\dfrac{1}{2},22\dfrac{3}{4},...\] is the first negative term?

Hint: The given question is of a sequence, we have to find the first negative term of the sequence.
Before finding the first negative term we have to find the type of the given series.
Using the first term and the common divisor we will find whether it is in AP.

Formula used: We know that, the nth term of arithmetic progression with first term as \[a\] and common divisor \[d\] is
\[a + (n - 1)d\]
Then we can find the first negative term.

Complete step-by-step answer:
It is given that; the sequence is \[25,24\dfrac{1}{4},23\dfrac{1}{2},22\dfrac{3}{4},...\]
We have to find the first negative term of this given sequence.
At first, we will find out the nature of the sequence.
The simplified form of the sequence is: \[25,\dfrac{{97}}{4},\dfrac{{47}}{2},\dfrac{{91}}{4},...\]
Now, we will convert the sequence in the same denominator. Hence, the sequence will be:
\[ \Rightarrow \dfrac{{100}}{4},\dfrac{{97}}{4},\dfrac{{94}}{4},\dfrac{{91}}{4},...\]
Here, \[100,97,94,91,...\] are in arithmetic progression.
So, \[\dfrac{{100}}{4},\dfrac{{97}}{4},\dfrac{{94}}{4},\dfrac{{91}}{4},...\] are in arithmetic progression.
Therefore, the first term \[a = \dfrac{{100}}{4}\], the common difference \[d = \dfrac{{97}}{4} - \dfrac{{100}}{4} = \dfrac{{94}}{4} - \dfrac{{97}}{4} = \dfrac{{ - 3}}{4}\]
We know that, the nth term of arithmetic progression with first term as \[a\] and common divisor \[d\] is
\[a + (n - 1)d\].
Now, as per the problem, the term should be negative.
So, we can write as:
\[ \Rightarrow a + (n - 1)d < 0\]
Substitute the value of first term and common divisor we get,
\[ \Rightarrow \dfrac{{100}}{4} + (n - 1)(\dfrac{{ - 3}}{4}) < 0\]
Simplifying we get,
\[ \Rightarrow 100 - 3n + 3 < 0\]
Simplifying again we get,
\[ \Rightarrow 3n > 103\]
Simplifying again we get,
\[ \Rightarrow n > \dfrac{{103}}{3}\]
Simplifying again we get,
\[ \Rightarrow n > 34\dfrac{1}{3}\]
So, the first negative term is \[{35^{th}}\] term.
Hence, \[{35^{th}}\] of the sequence \[25,24\dfrac{1}{4},23\dfrac{1}{2},22\dfrac{3}{4},...\] is the first negative term.

Note: Let us find the term.
We know that, the nth term of arithmetic progression with first term as \[a\] and common divisor \[d\] is
\[a + (n - 1)d\]
Substitute the value of first term and common divisor that is \[a = 25,d = \dfrac{{ - 3}}{4},n = 34\] we get,
\[ \Rightarrow 25 + (34 - 1)\dfrac{{ - 3}}{4} = \dfrac{1}{4}\]
Substitute the value of first term and common divisor that is \[a = 25,d = \dfrac{{ - 3}}{4},n = 35\] we get,
\[ \Rightarrow 25 + (35 - 1)\dfrac{{ - 3}}{4} = \dfrac{{ - 1}}{2}\]
Hence, \[{35^{th}}\] of the sequence \[25,24\dfrac{1}{4},23\dfrac{1}{2},22\dfrac{3}{4},...\] is the first negative term and that is \[\dfrac{{ - 1}}{2}\].