Question

Which term of the sequence $17,16\dfrac{1}{5},15\dfrac{2}{5},14\dfrac{3}{5},\ldots\$ is the first negative term?
A. $23^{\text{rd}}$
B. $3^{\text{rd}}$
C. $33^{\text{rd}}$
D. None of these

Answer 1

In this question, we need to find which term of the given sequence $17,16\dfrac{1}{5},15\dfrac{2}{5},14\dfrac{3}{5},\ldots\$ is the first negative term. On observing the given sequence, it is an arithmetic progression. It is nothing but a sequence where the difference between the consecutive terms are the same. From the given series, we can find the first term $(a)$ and the common difference $(d)$ . Then , by using the formula of arithmetic progression, we can easily find $n$. In order to find the first negative term, we need to suppose that the last non-negative term of the series is definitely close to zero, that is $a_{n} \simeq 0$ .

Formula used :
The Formula used to find the nth terms in arithmetic progression is
$a_{n} = \ a + \left( n- \ 1 \right)d$
Where $a$ is the first term
$d$ is the common difference
$n$ is the number of term
$a_{n}$ is the $n^{\text{th}}$ term

Complete answer:
Given, $17,16\dfrac{1}{5},15\dfrac{2}{5},14\dfrac{3}{5},\ldots\$
Here we need to find which term of the given sequence $17,16\dfrac{1}{5},15\dfrac{2}{5},14\dfrac{3}{5}\ldots$ is the first negative term.
Thus $a$ is $17$ and $d = \left( 16\dfrac{1}{5} – 17 \right)$
On simplifying,
We get,
$\Rightarrow \ d = \dfrac{81}{5} – 17 = \left( \dfrac{81 – 85}{5} \right)$
On further simplifying,
We get,
$d = - \dfrac{4}{5}$
Let us consider the first negative term of the series be $t_{n}$. Since it is a negative term, it is less than zero.
Now,
$\Rightarrow \ t_{n} = a + \left( n-\ 1 \right)d$
Since it is a negative term, it is less than zero.
That is $t_{n} < 0$
$\Rightarrow \ a + \left( n\ -\ 1 \right)d < 0$
On substituting the known values,
We get,
$17 + \left( n – 1 \right)\left( - \dfrac{4}{5} \right) < 0$
On simplifying,
We get,
$17 - \dfrac{4}{5}n + \dfrac{4}{5} < 0$
On further simplifying,
We get,
$\Rightarrow \ - \dfrac{4}{5}n + \dfrac{85 + 4}{5} < 0$
On removing the parentheses,
We get,
$- \dfrac{4}{5}n + \dfrac{89}{5} < 0$
On subtracting both sides by $\dfrac{89}{5}$,
We get,
$- \dfrac{4}{5}n < \- \dfrac{89}{5}$ [$( - )$ get cancelled, since it is in both sides ]
On multiplying both sides by $\dfrac{5}{4}$ ,
We get,
$\Rightarrow \ n < \dfrac{89}{5} \times \dfrac{5}{4}$
On simplifying,
We get,
$n < \dfrac{89}{4}$
Now on converting the fraction in the form of mixed fraction,
We get,
$n = 22\dfrac{1}{4}$
Therefore, the first negative term is the $23^{\text{rd}}$ term.
Final answer :
The first negative term is the $23^{\text{rd}}$ term.
Option A).$23^{\text{rd}}$ is the correct answer.

Therefore, the correct option is A

Note: In order to solve these types of questions, we need to observe that the first negative term of the series is less than zero. We also must remember that , we must round off n to its nearest whole number value to find the first negative term of the series. We should be very careful in choosing the correct formula because there is the chance of making mistakes in interchanging the formula of finding term and summation of term. If we try to solve this sum with the formula $S_{n} = \dfrac{n}{2}\left( a + l \right)$ where $l$ is the last term of the series , then our answer will be totally different and can get confused.