Question:medium

Numerically greatest term in the expansion of $(2x-3y)^n$ when $x=\frac{7}{5}, y=\frac{3}{7}$ and $n=13$ is

Show Hint

When calculating the numerically greatest term, be very careful with the formula for the ratio, which is $\frac{n-r+1}{r} |\frac{B}{A}|$. A common mistake is using $r+1$ in the denominator. If your calculated value does not match the options, double-check your arithmetic. If it still doesn't match, the question might be flawed, which can happen in competitive exams.
Updated On: Mar 26, 2026
  • $13.3^5.7^9$
  • $13.3^4.7^9$
  • $26.3^5.7^9$
  • $26.3^4.7^9$
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Formula for Numerically Greatest Term (N.G.T) For expansion \( (A+B)^n \), find \( m = \frac{(n+1)|B|}{|A|+|B|} \). Here \( A = 2x = 7 \), \( B = -3y = -3(3/7) = -9/7 \). \( |A| = 7, |B| = 9/7 \).
Step 2: Calculate m \[ m = \frac{(13+1) \cdot \frac{9}{7}}{7 + \frac{9}{7}} = \frac{14 \cdot \frac{9}{7}}{\frac{49+9}{7}} = \frac{18}{\frac{58}{7}} \times \frac{7}{7} \text{ (Wait, simplifying fraction)} \] \[ m = \frac{14 \cdot (9/7)}{58/7} = \frac{14 \cdot 9}{58} = \frac{126}{58} \approx 2.17 \] Since \( m \) is not an integer, the N.G.T is \( T_{[m]+1} \). \( [m] = 2 \), so the greatest term is \( T_{2+1} = T_3 \).
Step 3: Calculate \( T_3 \) \( T_{r+1} = \binom{n}{r} A^{n-r} B^r \). Here \( r=2 \). \( T_3 = \binom{13}{2} (2x)^{11} (-3y)^2 \). \( T_3 = \frac{13 \times 12}{2} (7)^{11} \left(\frac{9}{7}\right)^2 \). \( T_3 = 78 \cdot 7^{11} \cdot \frac{3^4}{7^2} = 78 \cdot 7^9 \cdot 3^4 \).
Step 4: Match with Options \( 78 = 2 \times 3 \times 13 = 26 \times 3 \). \( T_3 = (26 \times 3) \cdot 3^4 \cdot 7^9 = 26 \cdot 3^5 \cdot 7^9 \).
Was this answer helpful?
0