Q56 / 150 · MHT CET 2026 (19 April Shift 2) · Mathematics
If ${}^{n}C_4, {}^{n}C_5$ and ${}^{n}C_6$ are in arithmetic progression (A.P.), then the value of n is...
- A$5$ or $11$
- Correct answerB$7$ or $14$
- C$8$ or $15$
- D$6$ or $13$
Solution
Given that $^{n}C_4, ^{n}C_5, ^{n}C_6$ are in A.P.
$2 \times ^{n}C_5 = ^{n}C_4 + ^{n}C_6$
$2 \times \dfrac{n!}{5!(n-5)!} = \dfrac{n!}{4!(n-4)!} + \dfrac{n!}{6!(n-6)!}$
Multiplying the entire equation by $\dfrac{6!(n-4)!}{n!}$:
$2 \times 6 \times (n-4) = 6 \times 5 + (n-4)(n-5)$
$12n - 48 = 30 + n^2 - 9n + 20$
$n^2 - 21n + 98 = 0$
$(n-7)(n-14) = 0$
$n = 7$ or $n = 14$
Answer: $7$ or $14$
$2 \times ^{n}C_5 = ^{n}C_4 + ^{n}C_6$
$2 \times \dfrac{n!}{5!(n-5)!} = \dfrac{n!}{4!(n-4)!} + \dfrac{n!}{6!(n-6)!}$
Multiplying the entire equation by $\dfrac{6!(n-4)!}{n!}$:
$2 \times 6 \times (n-4) = 6 \times 5 + (n-4)(n-5)$
$12n - 48 = 30 + n^2 - 9n + 20$
$n^2 - 21n + 98 = 0$
$(n-7)(n-14) = 0$
$n = 7$ or $n = 14$
Answer: $7$ or $14$
Hence, the correct answer is option: 'B'.