JEE PYQ: Sequence And Series Question 67
Question 67 - 2019 (12 Jan Shift 1)
If $^nC_4$, $^nC_5$ and $^nC_6$ are in A.P., then $n$ can be:
(1) 9
(2) 14
(3) 11
(4) 12
Show Answer
Answer: (2)
Solution
Since $^nC_4$, $^nC_5$ and $^nC_6$ are in A.P.
$2 \cdot ^nC_5 = ^nC_4 + ^nC_6$
$2 = \frac{^nC_4}{^nC_5} + \frac{^nC_6}{^nC_5}$
$2 = \frac{5}{n-4} + \frac{n-5}{6}$ (wait, let me just solve)
$\Rightarrow n^2 - 21n + 98 = 0$
$(n - 7)(n - 14) = 0$
$n = 7, n = 14$
BETA
