If 1 + x^4 + x^3 + x^2 + x = 80, then the average (arithmetic mean) of

Question

If 1 + x^4 + x^3 + x^2 + x = 80, then the average (arithmetic mean) of x, x^2, x^3, x^4 and x^5 is equal to which of the following?

A. 12x
B. 13x
C. 14x
D. 16x
E. 20x

Bunuel · Accepted Answer

OFFICIAL SOLUTION:

The average (arithmetic mean) of x, x^2, x^3, x^4 and x^5 is (x + x^2 + x^3 + x^4 + x^5)/5 = x(1 + x + x^2 + x^3 + x^4)/5 = 80x/5 = 16x.

Answer: D.

askhere · Answer

1 + x^4 + x^3 + x^2 + x = 81
i.e. 1 +x+ x^2+ x^3+ x^4 = 81
 x+ x^2+ x^3+ x^4 = 80

x(1 +x+ x^2+ x^3) = 80
 x(81-x^4) = 80

81x - x^5 = 80
 x^5 = 81x -80

Now x+ x^2+ x^3+ x^4+ x^5 = 80 + 81x -80 = 81x

Average of {x, x^2, x^3, x^4, x^5} = 81x/5 ~ 16x

Answer  D

Ambarish

NavenRk · Answer

1 + x^4 + x^3 + x^2 + x = 81

multiplying by 'x' on both sides

x(1 + x^4 + x^3 + x^2 + x )= 81x

dividing by 5 we get

x+ x^2+ x^3+ x^4+x^5 = 81x/5 Approx 16x
But is approximation allowed? i don't see it mentioned in the question?

askhere · Answer

Ohh yeah. This is a much easier way. Kudos!
 I think the question meant approximate value.

D3N0 · Answer

Ans: D

Solution:  /5
Known 1 + x^4 + x^3 + x^2 + x = 81
taking x common from the first equation it becomes 
= *x/5
=81*x/5
=16x Answer
Ans.: D

ablatt4 · Answer

It is not possible to solve this by solving for x, don't bother.

If we need the average of x, x^2, x^3, x^4 and x^5, we can multiply our current system by x and get x+x^2+x^3+ x^4+x^5=80x

if the average is 5 terms going into 80x then 80x/5=16x D

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