Question

If the numbers $a,b,c,d$ and $e$ are in the form of an AP with $a=1$, then $a-4b+6c-4d+e$ is equal to

• A. $1$
• B. $2$
• C. $0$
• D. $3$

Answer 1

The correct option is C

$0$

Explanation for correct option:

Given that the numbers $a,b,c,d$ and $e$ are in AP. and $a=1$

Let the common difference be $h$

$$\begin{gathered} \therefore b=a+h \\ =1+h \\ \therefore c=a+2h \\ =1+2h \\ \therefore d=a+3h \\ =1+3h \\ \therefore e=a+4h \\ =1+4h \end{gathered}$$

$$\begin{gathered} \therefore a-4b+6c-4d+e \\ =1-4\left(1+h\right)+6\left(1+2h\right)-4\left(1+3h\right)+1+4h \\ =1-4-4h+6+12h-4-12h+1+4h \\ =0 \end{gathered}$$

Hence, option C is correct answer.