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8. Sequences and Series
easy
If $a,\;b,\;c,\;d,\;e,\;f$ are in $A.P.$, then the value of $e - c$ will be
A
$2(c - a)$
B
$2(f - d)$
C
$2(d - c)$
D
$d - c$
Solution
(c) $a,\;b,\;c,\;d,\;e,\;f$ are in $A.P.$
So $b – a = c – b = d – c = e – d = f – e = K$
Where $K$ is a common difference.
Now, $d – c = e – d$
$ \Rightarrow $$e + c = 2d$.
$e{\rm{-}}c + {\rm{2}}c = 2d $
$\Rightarrow e – c = 2(d – c)$.
Trick : Check by putting $a = 1,\;b = 2,\;c = 3,\;d = 4,\;e = 5$ and $f = 6$.
Standard 11
Mathematics
