Question

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

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

Answer 1

The correct option is C $0$

Value of $a-4b+6c-4d+e=0$
1st term $=A=a$
2nd term $=A+D=b$
3rd term $=A+2D=c$
4th term $=A+3D=d$
5th term $=A+4D=e$
$\therefore a-4b+6c-4d+e=A-4(A+D)+6(A+2D)-4(A+3D)+(A+4D)$

$=A-4A-4D+6A+12D-4A-12D+A+4D$

$=0$