MAH-CET 2026 — Mathematics PYQ

Step 1: Write the formula for the general term

In the binomial expansion of $(x+y{)}^n$, the general term $T_{r+1}$ is given by the formula:

$$T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)x^{n-r}{y}^r$$

For the given expression $(a-b{)}^n$, we can rewrite it as $[a+(-b){]}^n$. Thus, the general term formula becomes:

$$T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)a^{n-r}(-b{)}^r$$

Step 2: Find the expressions for the $5^{\text{th}}$ and $6^{\text{th}}$ terms

• For the $5^{\text{th}}$ term ($T_5$), set $r=4$:

  $$T_5=\left(\genfrac{}{}{0ex}{}{n}{4}\right)a^{n-4}(-b{)}^4=(\genfrac{}{}{0ex}{}{n}{4})a^{n-4}{b}^4$$

• For the $6^{\text{th}}$ term ($T_6$), set $r=5$:

  $$T_6=\left(\genfrac{}{}{0ex}{}{n}{5}\right)a^{n-5}(-b{)}^5=-(\genfrac{}{}{0ex}{}{n}{5})a^{n-5}{b}^5$$

Step 3: Apply the given condition ($T_5+T_6=0$)

According to the problem statement:

$$T_5+T_6=0⟹T_5=-T_6$$

Substitute the evaluated terms into this relation:

$$\left(\genfrac{}{}{0ex}{}{n}{4}\right)a^{n-4}{b}^4=-\left[-\left(\genfrac{}{}{0ex}{}{n}{5}\right)a^{n-5}{b}^5\right]$$

$$\left(\genfrac{}{}{0ex}{}{n}{4}\right)a^{n-4}{b}^4=\left(\genfrac{}{}{0ex}{}{n}{5}\right)a^{n-5}{b}^5$$

Step 4: Solve for $\frac{a}{b}$

Rearrange the equation to isolate the variables $a$ and $b$ on one side:

$$\frac{a^{n-4}}{a^{n-5}}\cdot \frac{b^4}{b^5}=\frac{\left(\genfrac{}{}{0ex}{}{n}{5}\right)}{\left(\genfrac{}{}{0ex}{}{n}{4}\right)}$$

Using exponent laws, $\frac{a^{n-4}}{a^{n-5}}=a^{(n-4)-(n-5)}=a^1$, and $\frac{b^4}{b^5}=\frac{1}{b}$:

$$\frac{a}{b}=\frac{\left(\genfrac{}{}{0ex}{}{n}{5}\right)}{\left(\genfrac{}{}{0ex}{}{n}{4}\right)}$$

Now, expand the combinations using the definition $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{r!(n-r)!}$:

$$\frac{a}{b}=\frac{\frac{n!}{5!(n-5)!}}{\frac{n!}{4!(n-4)!}}$$

Cancel out $n!$ from both the numerator and denominator:

$$\frac{a}{b}=\frac{4!(n-4)!}{5!(n-5)!}$$

Simplify the factorials ($5!=5\times 4!$ and $(n-4)!=(n-4)\times (n-5)!$):

$$\frac{a}{b}=\frac{4!\cdot (n-4)(n-5)!}{5\cdot 4!\cdot (n-5)!}$$

$$\frac{a}{b}=\frac{n-4}{5}$$

Correct Answer

(b) $\frac{n-4}{5}$