Question

which term of the sequence 23,22 1/2 ,22 , 21 1/2 ...... is the fist -ve term

Answer 1

$\underline{\underline{\mathfrak{\Large{Solution : }}}}$



$\textsf{Let , its nth term is first -ve term.}$



$\underline{\textsf{Given,}} \\ \\ \sf \implies First \: term (a) \: = \: 23 \\ \\ \sf \implies Common \: difference (d) \: = \: 22 \dfrac{1}{2} \: - \: 23 \\ \\ \qquad \qquad \qquad \qquad \qquad \qquad \: \sf = \: \dfrac{45}{2} \: - \: 23 \\ \\ \qquad \qquad \qquad \qquad \qquad \qquad \: \sf = \: \dfrac{45 \: - \: 46}{2} \: = \: - \dfrac{1}{2}$




$\underline{\textsf{Now,}} \\ \\ \sf \implies t_n \: < \: 0 \\ \\ \sf \implies a \: + \: ( n \: - \: 1 )d \: < \: 0 \\ \\ \sf \implies 23 \: + \: (n \: - \: 1 )(-0.5) \: < \: 0 \\ \\ \sf \implies23 \: - \: 0.5n \: + \: 0.5 \: < \: 0 \\ \\ \sf \implies23.5 \: - \: 0.5n \: < \: 0 \\ \\ \sf \implies - 0.5n \: < \: - 23.5 \\ \\ \sf \implies{n} \: > \: \dfrac{ - 23.5}{ - 0.5} \\ \\ \sf \: \: \therefore \: \: n \: > \: 47$




$\textsf{All the terms greater than 47th terms will be } \\ \textsf{negative. Since, we have to find the first negative } \\ \textsf{term , so the term will be just greater than 47.} \\ \textsf{Hence, the first negative term is 48th term.}$

Answer 2

Answer:

48th term

Step-by-step explanation:

We do this by assuming the nth term to be 0

In this AP,

First term = a = 23

$a_{n}$ = 0

Common difference = d = $a_{2} - a_{1}$

$a_{2} = 22\frac{1}{2} = \frac{44 + 1}{2} = \frac{45}{2}$

$a_{2} - a_{1} = \frac{45}{2}- 23$

= $\frac{45 - 46}{2}$

= $\frac{-1}{2}$

n =?

$a_{n}$ = a + (n - 1) d = 0

23 + (n - 1) $\frac{-1}{2}$ = 0

23 + $(\frac{-n + 1}{2})$ = 0

$\frac{46 - n + 1}{2}$ = 0

47 - n = 0

- n = -47

n = 47

Since the 47th term is zero, the first negative term is the 48th term