Given AP is 21 ,18,15,...
First term = 21
Common difference = 18-21 = -3
Let an = -81
We know that
an = a+(n-1)d
⇛21+(n-1)(-3) = -81
⇛ 21-3n+3 = -81
⇛24-3n = -81
⇛ 24+81 = 3n
⇛ 105= 3n
⇛ n = 105/3
⇛ n = 35
35th term of the AP is -81.
<u>also</u><u> read</u><u> similar</u><u> questions</u><u>:</u> Which term of a AP 5 , 13 , 21 ,... is 181?
brainly.com/question/2406241?referrer
Which term of the AP:3, 8, 13, 18,...,is 78?
brainly.com/question/15380012?referrer
Is 16 an option because if it is then... it's right?
<u>Answer-</u>
<em>Area of the triangle is </em><em>10 sq.units</em>
<u>Solution-</u>
We know that,
![\text{Area of the triangle}=\dfrac{1}{2}[{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]](https://tex.z-dn.net/?f=%5Ctext%7BArea%20of%20the%20triangle%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5B%7Bx_1%28y_2-y_3%29%2Bx_2%28y_3-y_1%29%2Bx_3%28y_1-y_2%29%5D)
Taking,
(x₁, y₁) = (3, 3)
(x₂, y₂) = (3, -1)
(x₃, y₃) = (-2, -5)
Then putting these in the formula,
![\text{Area of the triangle}=\dfrac{1}{2}[3(-1+5)+3(-5-3)-2(3+1)]](https://tex.z-dn.net/?f=%5Ctext%7BArea%20of%20the%20triangle%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5B3%28-1%2B5%29%2B3%28-5-3%29-2%283%2B1%29%5D)
![=\dfrac{1}{2}[3(4)+3(-8)-2(4)]](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B1%7D%7B2%7D%5B3%284%29%2B3%28-8%29-2%284%29%5D)
![=\dfrac{1}{2}[12-24-8]](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B1%7D%7B2%7D%5B12-24-8%5D)
![=\dfrac{1}{2}[-20]](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B1%7D%7B2%7D%5B-20%5D)
As area can not be negative, ignoring negative sign,
