Question

Which term of the $A.P,19,18\dfrac{1}{5},17\dfrac{2}{5},.....$ is the first negative term?
A. 24
B. 25
C. 26
D. 23

Answer 1

Hint: We will first start by using the fact that the ${{n}^{th}}$ term of an A.P with first term a and common difference d is $a+\left( n-1 \right)d$. Then we will use the fact that for the term to be negative it should be less than zero. Using this and general term we will find the term which is negative.

Complete step-by-step solution -
Now, we have been given an A.P as,
$19,18\dfrac{1}{5},17\dfrac{2}{5},.....$
Now, we have the first term of this A.P as a = 19.
Now, we have the common difference of the,
$\begin{align}
  & A.P=d=18\dfrac{1}{5}-19 \\
 & =\dfrac{91}{5}-19 \\
 & =\dfrac{91-95}{5} \\
 & =\dfrac{-4}{5} \\
\end{align}$
Now, we know that the ${{n}^{th}}$ term of the A.P is $a+\left( n-1 \right)d$. So, we have for this A.P,
${{T}_{n}}=19+\left( n-1 \right)\left( \dfrac{-4}{5} \right)$
Now, for first negative we will find the smallest n which satisfy ${{T}_{n}} < 0$. So, we have,
$
 \Rightarrow 19+\left( n-1 \right)\left( \dfrac{-4}{5} \right) < 0 \\
 \Rightarrow 19-\dfrac{4}{5}\left( n-1 \right) < 0 \\
 \Rightarrow 19 < \dfrac{4}{5}\left( n-1 \right) \\
 \Rightarrow \dfrac{95}{4}+1 < n \\
  \Rightarrow n > \dfrac{95+4}{4} \\
 \Rightarrow n > \dfrac{99}{4} \\
$
So, we have the smallest integral n which satisfies this as n = 25.
Hence, the correct option is (B).

Note: It is important to note that we have first found the ${{n}^{th}}$ term of the A.P and then use the fact that for negative term ${{T}_{n}} < 0$. So, we have to find the solution of this inequality but since we have to find the first negative term we have taken the smallest n which satisfies ${{T}_{n}} < 0$.