Question

If the numbers $a,b,c,d,e$ form an A.P with $a=1$, then find the value of $a-4b+6c-4d+e$.
A. 1
B. 2
C. 0
D. 3

Answer 1

Hint: The given variables are in an arithmetic progression so there will be a common difference and nth term can be calculated using suitable formulas.

Formula Used:
The general formula of arithmetic progression of nth term is $t_n=a+(n-1)d$, where $a$ is the first term, $n$ is the term number and $d$ is the common difference.

Complete step by step solution:
Given that $a,b,c,d,e$ are in the arithmetic progression.
So, the first term of the arithmetic progression is a
$t_1=a=1$
Assuming the common difference is $x$ . Then, using the nth term formula, rewrite each term as follows:
$t_n=a+(n-1)d$
The value of second term is
$t_2=1+(2-1)x$
$\Rightarrow b=1+x$
The value of third term is
$t_3=1+(3-1)x$
$\Rightarrow c=1+2x$
The value of fourth term is
$t_4=1+(4-1)x$
$\Rightarrow d=1+3x$
The value of fifth term is
$t_5=1+(5-1)x$
$\Rightarrow e=1+4x$
Next putting the above values in the given expression $a-4b+6c-4d+e$ , we get
$a-4b+6c-4d+e$
Substitute the values of $a,b,c,d,e$
$=1-4(1+x)+6(1+2x)-4(1+3x)+1+4x$
Simplifying this expression, we get
$=1-4-4x+6+12x-4-12x+1+4x$
Simplifying this expression, we get
$=6-6=0$

Option ‘C’ is correct

Note: Remember the properties and related formulas for an Arithmetic Progression. Also, we need to take care of multiplication of signs.
$(+)\times (-)=(-)$
$(-)\times (-)=(+)$
$(-)\times (+)=(-)$