Which term of sequence $4,9,14,19.........$ is 124

Hint: Find common difference of sequence
 common difference $={{T}_{r+1}}-{{T}_{r}}$ , if the difference between two consecutive terms is constant then the given sequence is in A.P.
 Apply ${{T}_{n}}=a+(n-1)(d)$

Complete step-by-step answer:
We have the sequence as from question
$4,9,14,19........$
As we see here that difference between two consecutive terms is
$9-4=14-9=19-4=5$
i.e. difference between two consecutive terms is constant so the given sequence is of A.P
Let us assume that the first term of an A.P. is ‘a’ and common difference is ‘d’. so, as we know nth term of an AP is given by
${{t}_{r}}=a+(r-1)(d)$
Now we have${{t}_{r}}=124$ and from question
$\begin{align}
  & a=4 \\
 & d=5 \\
\end{align}$
And we have to find the value of r so putting all the values we can write
$124=4+(n-1)5$
So, transposing 4 to L.H.S.
$124-4=(n-1)5$
Dividing both sides by 5
$\dfrac{120}{5}=n-1$
Adding both sides 1 we get
$\begin{align}
  & =n-1+1=24+1 \\
 & \Rightarrow n=25 \\
\end{align}$
Total number of term is 25
Hence option A is correct

Note: If common difference i.e. differences between two consecutive terms is constant then the given sequence is in AP. If common difference is +ve then sequence is in increasing order and if it is –ve the sequence is in decreasing order. Sequence is written after using comma, when we place + sign instead of comma we get series.