Question

A person purchases a television set for Rs. 16000. Its life is estimated to be 20 yaers. If the yearly depreciation is assumed to be constant, find the rate of depreciation and the price after 8 years.(Arithmetic Progression)

Please answer fast.

Answer 1

here, ATQ,
a = 16000
d = -16000/20 = -800

so,
a8 = a+7d = 16000+7(-800) = 16000-5600 = 10400

so,
rate of deprecation = 800
price after 8 years = 10400

Answer 2

Answer:

Step-by-step explanation:

Now since the life of the tv is 20 yrs, therefore its value would evaluate to be zero after 20 yrs.

which implies in 21st yr the tv will amount to 0.

we know, Tn= a + ( n - 1 )d

Here; n = 21; a = Rs 16000; d = -d ( since it's reducing with time (depreciating)):

Therefore, T 21 = 16000 + (21 - 1) (-d)

= 0 ( since T21 amounts to zero) = 16000 - 20d

= 20d = 16000

= d = 16000/20 = Rs 800

therefore rate of depreciation = Rs 800 per yr.

Now, taking n =9 ( since the price of the tvafter 8 yrs is asked; a = 16000; d = -d = -800 and putting it in the formula of nth term

T9 = 16000 + (9-1)(-800)

= 16000+ 8*(-800)

= 16000 - 6400

= 9600

Therefore the price of the TV after 8 yrs = Rs 9600 ans.