Gujarati
8. Sequences and Series
easy

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

Solution

(c) Let $D$ be the common difference of the $A.P.$

Then, $a – 4b + 6c – 4d + e  $

$=a – 4(a + D) + 6(a + 2D) – 4(a + 3D) + (a + 4D) = 0$.

Standard 11
Mathematics

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