Question

Which term of the progression -4/3 , -1 , -2/3....., -4 1/3 is the first positive term​

Answer 1

$\textbf{Given:}$

$\text{A.P is \frac{-4}{3},-1,\frac{-2}{3}........ }$

$\textbf{To find:}$

$\text{First positive term}$

$\textbf{Solution:}$

$\text{Consider A.P \frac{-4}{3},-1,\frac{-2}{3}........ }$

$d=-1+\frac{4}{3}=\frac{1}{3}$

$a=\dfrac{-4}{3}$

$\text{Let t_n be the first positive term}$

$\text{Then, t_n\,>\,0 }$

$\implies\,a+(n-1)d\,>\,0$

$\implies\,\dfrac{-4}{3}+(n-1)\dfrac{1}{3}\,>\,0$

$\implies\,\dfrac{-4+n-1}{3}\,>\,0$

$\implies\,\dfrac{n-5}{3}\,>\,0$

$\implies\,n-5\,>\,0$

$\implies\,n\,>\,5$

$\text{n is a positive integer just greater than 5}$

$\therefore\,\bf\,n=6$

$\textbf{Hence 6 th term is the first positive term}$

Find more:

29) Find the 31s term of an A.P., whose 11th term is 88 and 16th term is 73. Which

term of this series will be the 1st negative term?​

https://brainly.in/question/15897275

Answer 2

The 6th term of the progression is the first positive term.

Step-by-step explanation:

Given, $\frac{-4}{3}, -1, \frac{-2}{3},......$

1st term = $\frac{-4}{3}$

Common difference = d = $-1 - (\frac{-4}{3}) = \frac{1}{3}$

2nd term = $\frac{-4}{3} + \frac{1}{3} = \frac{-3}{3} = -1$

3rd term = $-1 +\frac{1}{3} = \frac{-2}{3}$

4th term = $\frac{-2}{3} + \frac{1}{3} = \frac{-1}{3}$

5th term = $\frac{-1}{3} + \frac{1}{3} = 0$

6th term = $0 + \frac{1}{3} = \frac{1}{3}$

Hence, the 6th term of the progression is the first positive term.