If a,;b,;c,;d,;e,;f are in A.P., then the val | vedclass

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(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 -

Vedclass · Accepted Answer

$2(d - c)$

Vedclass · Answer

(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 $f = 6$.

If $a,\;b,\;c,\;d,\;e,\;f$ are in $A.P.$, then the value of $e - c$ will be

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