Question

find the first negative term of an ap with sequence 20,19 1/2..

Answer 1

$\textsf{Consider the AP \ 20,\ 19 \frac{1}{2},\ 19,... } \\ \\ \\ a=20 \\ \\ d = 19 \frac{1}{2} - 20 = - \frac{1}{2} \\ \\ \\ a_n = dn + a - d \\ \\ a_n = - \frac{1}{2} n + 20 - (- \frac{1}{2}) \\ \\ a_n = - \frac{1}{2} n + 20 + \frac{1}{2} \\ \\ a_n = 20 \frac{1}{2} - \frac{1}{2} n \\ \\ a_n = \frac{1}{2} (41 - n)$

$\textsf{From the algebraic form of the AP, we can make out that the terms will be}\\ \textsf{positive if n<41 and will be negative if n>41 .} \\ \\ \\ \textsf{If n<41,\ 41-n will be positive and so will be a_n = \frac{1}{2}(41-n) .} \\ \\ \textsf{If n>41,\ 41-n will be negative and so will be a_n = \frac{1}{2}(41-n) .}$

$\textsf{As n is the no. of terms of the AP or the position number of a particular term}\\ \textsf{in the AP which is always a positive integer, the lowest value of n for the term of the AP}\\ \textsf{being negative is 42 .} \\ \\ \textsf{This means that 42^{nd} term of the AP is the first negative term. } \\ \\ \\ \textsf{Thus,}$

$a_{42}=\frac{1}{2}(41-42) \\ \\ a_{42}=\frac{1}{2} \times -1 \\ \\ a_{42}=\bold{-\frac{1}{2}}$

$\textsf{Thus \bold{-\frac{1}{2}} is the first negative term of the AP. }$

Answer 2

Answer:

Step-by-step explanation:

In given AP   a= 20    and   d = -1/2  

so clearly  all integers less than 20 are also the terms

so    0   is a term

o + d  is next term of 0  

0 + d = 0 -1/2 = -1/2   is first negative term