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8. Sequences and Series
easy
If $a,\,b,\,c$ are in $A.P.$, then $(a + 2b - c)$ $(2b + c - a)$ $(c + a - b)$ equals
A
$\frac{1}{2}abc$
B
$abc$
C
$2\ abc$
D
$4\ abc$
Solution
(d) $(a + 2b – c)\,(2b + c – a)\,(c + a – b)$
$ = (a + a + c – c)(a + c + c – a)(2b – b)$ $ = 4\,abc.$
$(\because a,b,c$ are in $A.P.$,
$\therefore 2b = a + c)$.
Standard 11
Mathematics