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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