Question

if a b c d e are the AP then find the values of a-4b+6c-4d+e

Answer 1

Given that a,b,c,d,e are in Arithmetic progression.

that means first term a1=a

let common difference be D then we can rewrite each term using nth term formula as:

an=a1+(n-1)d

a

b=a+x

c=a+2x

d=a+3x

e=a+4x

Now plug these values into given expression a-4b+6c-4d+e

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

=a-4(a+x)+6(a+2x)-4(a+3x)+(a+4x)

=a-4a-4x+6a+12x-4a-12x+a+4x

=a-4a+6a-4a+a-4x+12x-12x+4x

=(1-4+6-4+1)+(-4+12-12+4)x

=0+0x

=0+0

=0

Hence final answer is 0.

Answer 2

A.P. : a,b,c,d,e

--> Common Difference = D = b-a = c-b = d-c = e-d

By equating them one by one we get,

2b = a+c ; 2c = b+d ; 2d = c+e

Now,

a - 4b + 6c - 4d + e = a - 2(a+c) +6c - 2(c+e) + e

= a - 2a - 2c + 6c - 2c - 2e +e

= - a + 2c - e

= -a + b + d - e

= (b-a) - (e-d)

= D - D

= 0

So, a - 4b + 6c - 4d + e = 0