Solve the AP

Algebra Level 2

a,b,c,d,e are in AP. Then, the value of a-4b+6c-4d+e is

2 0 1 -1

2 solutions

Mahdi Raza
Jun 17, 2020

Let the numbers in the A.P. be as such with $c$ as the middle term and $x$ as the common difference: $(c-2x) (c-x), (c), (c+x), (c+2x)$

Then the expression will look like: $\begin{aligned} &= (c-2x) -4(c-x) + 6(c) -4(c+x) + (c+2x) \\ &= (c -4c + 6c -4c + c) + (-2x + 4x - 4x + 2x) \\ &= (\cancel{c -4c + 6c -4c + c}) + (\cancel{-2x + 4x - 4x + 2x}) \\ &= \boxed{0} \end{aligned}$

Shriniketan Ruppa
Jun 17, 2020

Let a,b,c,d,e be 1,2,3,4,5 respectively. We just gotta make sure that that the numbers are in AP. For example, a,b,c,d,e could also be 2,4,6,8,10. To make things easier I took it as 1,2,3,4, and 5.

Substituting the values we get,

=a-4b+6c-4d+e

=1-4(2)+6(3)-4(4)+5

=1-8+18-16+5

=24-24

@Shriniketan Ruppa : I got the question right, but I guessed the answer because I didn't know what AP is. What's AP? Is it like the same thing as an Arithmetic Sequence?

Ved Pradhan - 11 months, 4 weeks ago

@Ved Pradhan : AP is nothing but Arithmetic Progression.

Shriniketan Ruppa - 11 months, 4 weeks ago

@Shriniketan Ruppa : Oh, okay, that makes sense. I learned them as Arithmetic Sequences, not Arithmetic Progressions, but I guess they mean the same thing.

Ved Pradhan - 11 months, 4 weeks ago

Solve the AP

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