Question

a, b, c, d, e are in A.P., then the value of a – 4b + 6c – 4d + e is​

Answer 1

$\huge{\mathcal{\underline{\green{SOLUTION}}}}$

Let k be the Common Difference

Since a, b, c, d, e are in AP

SO

b = a+k

c = a+2k

d = a+3k

e = a+4k

Now

$\displaystyle \: a – 4b + 6c – 4d + e$

$= \displaystyle \: a – 4(a +k ) + 6(a + 2k) – 4( a+3k )+ ( a+4k )$

$= \displaystyle \: a – 4 a- 4k + 6a + 12k - 4a- 12k + a + 4k$

$= 0$

$\displaystyle\textcolor{red}{Please \: Mark \: it \: Brainliest}$

Answer 2

Answer: 0.

Given:

$\sf a,b,c,d\:and\:e\:are\:in\:A.P.$

To find:

$\sf The\:value\:of\:a-4b+6c-4d+e.$

Formula used:

$\boxed{\sf a_n=a+(n-1)d}$

Explanation:

$\sf Let\:A\:be\:the\:first\:term.\:Also, D\:be\:the\:common\:difference.$

$\sf\therefore a=A$

   $\sf\ b=A+D$

   $\sf c=A+2D$

   $\sf d=A+3D$

   $\sf e=A+4D$

$\sf Substituting\:the\:values\:of\:a,b,c,d\:and\:in\:the\:question,we\:get:$

$\Rightarrow\sf A-4(A+D)+6(A+2D)-4(A+3D)+(A+4D)$

$\Rightarrow\sf A-(4A+4D)+(6A+12D)-(4A+12D)+(A+4D)$

$\Rightarrow\sf A-4A-4D+6A+12D-4A-12D+A+4D$

$\Rightarrow\sf A-4A+6A-4A+A-4D+12D-12D+4D$  

$\sf\Rightarrow \boxed{\sf 0}$

Hence, value of a – 4b + 6c – 4d + e is​ 0.