QJEEM24Ma32a1: JEE Main 2024 Mathematics - Class 11 JEEM24Ma32a1 Solution: Let 0 ≤ r ≤ n . If n + 1 C r + 1 : n C r : n − 1 C r − 1 = 55 : …

Given $0 \leq r \leq n$ . If $\binom{n+1}{r+1} : \binom{n}{r} : \binom{n-1}{r-1} = 55 : 35 : 21$ , then we are to determine the value of $2n + 5r$ .

Step-by-Step Solution:

1. Write the given proportions involving binomial coefficients:

$\frac{n+1}{r+1} \times \binom{n}{r} : \binom{n}{r} : \frac{r}{n} \times \binom{n}{r} = 55 : 35 : 21$

1. Simplify the proportions:

$\frac{n+1}{r+1} = \frac{55}{35} \quad \text{and} \quad \frac{n}{r} = \frac{35}{21}$

1. From the simplified ratios, we establish the following two equations:

$\frac{n+1}{r+1} = \frac{11}{7} \quad \Rightarrow \quad 7(n+1) = 11(r+1) \quad \Rightarrow \quad 7n - 11r = 4 \quad \text{.... (1)}$

$\frac{n}{r} = \frac{5}{3} \quad \Rightarrow \quad 3n = 5r \quad \Rightarrow \quad 3n - 5r = 0 \quad \text{.... (2)}$

1. Solve equations (1) and (2) simultaneously:

• From equation (2), solve for $n$ :

$3n = 5r \quad \Rightarrow \quad n = \frac{5r}{3}$

• Substitute $n$ into equation (1):

$7 \left(\frac{5r}{3}\right) - 11r = 4 \quad \Rightarrow \quad \frac{35r}{3} - 11r = 4$

$\frac{35r - 33r}{3} = 4 \quad \Rightarrow \quad \frac{2r}{3} = 4 \quad \Rightarrow \quad 2r = 12 \quad \Rightarrow \quad r = 6$

• Substituting $r$ back to find $n$ :

$n = \frac{5 \times 6}{3} = 10$

1. Compute $2n + 5r$ :

$2n + 5r = 2 \times 10 + 5 \times 6 = 20 + 30 = 50$

Thus, the value of $2n + 5r$ is 50.