Question

Which term of the sequence 114,109,104.... is first negative term?

Answer 1

${\red{\huge{\underline{\mathbb{Given:-}}}}}$

AP : 114 , 109 , 104 ,..........

Here, a = 114 where a is the first term

d = 109 - 114 = -5 is the common difference

$\\ \huge{\bold { Solution}} \\ \\ We \: know ,\\ \: an < 0 \\ \\ So, \: nth \: term \: of \: ap \: is \: \\ \\ an \: = a + (n - 1)d \\ \\ \rightarrow a + (n - 1)d < 0 \\ \\ \rightarrow114 + (n - 1)( - 5) < 0 \\ \\ \rightarrow114 - 5n + 5 < 0 \\ \\ \rightarrow119 - 5n < 0 \\ \\ \rightarrow 119 < 5n \\ \\ \rightarrow n > \frac{119}{5} \\ \\ \rightarrow n > 23.8 \\ \\ n = 24 \\ \\ {\purple{\boxed{\large{\bold{The \: first \: negative \: term \: is \: 24th \: term}}}}}$

Hope its helpful ❤️

Answer 2

Question :

Which term of the sequence 114,109,104.... is first negative term?

Theory :

${\purple{\boxed{\large{\bold{a_{n}=a + (n - 1)d}}}}}$

Solution :

Given Ap series: 114,109,104....

In this series ;

first term ,a = 114

common difference ,d = 109-114 =-5

_________________________

we have to find first negative term in the given AP series .

⇒ The value of a+(n-1) d < 0

a+(n-1)d < 0

⇒ 114 + (n-1) -5 < 0

⇒114-5n +5 <0

⇒ 119-5n <0

⇒5n >119

⇒$n > \frac{119}{5}$

⇒n >23.8

as n is a integer, the first value which satisfies the above condition is n = 24

Now,

$a _{24} = a + (24- 1)d$

⇒ $a _{24} = 114+ (24- 1)-5$

⇒$a _{24} = 114+(23)-5$

⇒$a _{24} = 114-115=-1$

__________________________

Therefore first negative no is -1