Question

which term of the AP: 121, 117, 113, ..., is it's first negative term? ( HINT: find n and an< 0).

Answer 1

The given A.P is 121, 117, 113…

Here, first term a = 121 and common difference, d = –4.

Let n th term be the first negative term of the given A.P.

∴ an < 0

⇒ 121 + (n – 1) (–4) < 0  (Since, an = a + (n – 1) d)

⇒ –4n + 125 < 0

⇒ – 4n < – 125



∴ n = 32  [Since, n is a natural number]

Thus, 32nd term is the first negative term of the given A.P. HOPE THIS HELPS ...

Answer 2

$\bf\huge\boxed{\boxed{\bf\huge\:Hello\:Mate}}}$

$\bf\huge AP = 121 , 117 \:and\: 113$

$\bf\huge a = 121 , d => 117 - 121 = -4$

$\bf\huge => a_{n} = a + (n - 1)d$

$\bf\huge => 121 - 4n + 4 = 125 - 4n$

$\bf\huge First\: Negative\: Term$

$\bf\huge = a_{n} < 0$

$\bf\huge = 125 - 4 < 0$

$\bf\huge = 125 < 4n$

$\bf\huge = \frac{125}{4} < n$

$\bf\huge = 31 \frac{1}{4} <n$

$\bf\huge n \:is\: an\:integer\:and\:n >31 \frac{1}{4}$

$\bf\huge The\:first\:Negative\:Term\:=\:32nd\:term$

$\bf\huge\boxed{\boxed{\:Regards=\:Yash\:Raj}}}$