Question

Which term of the AP: 121, 117, 113,..., is its first negative term?
​

Answer 1

Answer:

The 32ⁿᵈ term of the AP is its first negative term.

Step-by-step-explanation:

The given AP is 121, 117, 113, ....

a = t₁ = 121

d = t₂ - t₁ = 117 - 121 = - 4

We have to find the first negative term of the A.P.

Let the first negative term be tₙ.

As the nᵗʰ term ( tₙ ) is negative, it's less than zero.

∴ tₙ < 0

⇒ a + ( n - 1 ) d < 0

⇒ 121 + ( n - 1 ) * - 4 < 0

⇒ 121 - 4n + 4 < 0

⇒ 121 + 4 - 4n < 0

⇒ 125 - 4n < 0

⇒ 125 < 4n

⇒ 125 ÷ 4 < n

⇒ 31.25 < n

∴ n = 32

∴ The 32ⁿᵈ term of the AP is its first negative term.

Answer 2

The given AP Is 121, 117, 113,..

$a = t_{1} = 121 \\ d = t_{2} - t_{1} \\ = 117 - 121 \\ = - 4$

We have to find the first negative term of the A.P.

$let \: \: the \: \: first \: \: negative \: \\ term \: \: be \: \: t_{n} \\ \\ as \: the \: {n}^{th} \: term \: be \: \: ( t_{n} )\: is \: \\ negative \: it \: is \: less \: than \: zero </p><p>t_{n} < 0$

= a + (n-1) d < 0

=121 + (n-1 )* - 4 < 0

=121 - 4n + 4 < 0

=121 + 4 - 4n < 0

=125 - 4n < 0

=125 - 4n

=125 ÷ 4 < n

=31.25 < n

n = 32

The 32nd term of the AP is its first negative term..