Question

If the numbers a, b, c, d, e form an A.P., then the value of $a-4b+6c-4d+e$ is?

• A. $1$
• B. $2$
• C. $0$
• D. None of these

Answer 1

Answer: (C)

$0$

Given that $a,b,c,d,e$ are in arithmetic progression,

That means first term $a1=1$

Let common difference be $D$ then using

$a_n=a_1+(n-1)d$ we can write

$a$

$b=a+x$

$c=a+2x$

$d=a+3x$

$l=a+4x$

Now

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

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

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

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

$=0+0x$

$=0$