MAH-CET 2026 — Mathematics PYQ

Let us solve this problem step-by-step using standard formulas for an Arithmetic Progression (AP).

Let the first term of the AP be $a$ and the common difference be $d$.

The formulas for the sum of $n$ terms ($S_n$) and the $n$-th term ($T_n$) are:

• $S_n=\frac{n}{2}[2a+(n-1)d]$

• $T_n=a+(n-1)d$

Step 1: Set up the ratio equation using the given conditions

We are given:

$$\frac{S_p}{S_q}=\frac{p^2}{q^2}$$

Expand $S_p$ and $S_q$ using the sum formula:

$$\frac{\frac{p}{2}[2a+(p-1)d]}{\frac{q}{2}[2a+(q-1)d]}=\frac{p^2}{q^2}$$

We can simplify the left side by canceling out the $\frac{1}{2}$ terms:

$$\frac{p[2a+(p-1)d]}{q[2a+(q-1)d]}=\frac{p^2}{q^2}$$

Now, divide both sides by $\frac{p}{q}$ (since $p\ne q$ and they are non-zero indexes):

$$\frac{2a+(p-1)d}{2a+(q-1)d}=\frac{p}{q}$$

Step 2: Find the mathematical relationship using a substitution shortcut

We need to find the value of the term ratio:

$$\frac{T_6}{T_{21}}=\frac{a+(6-1)d}{a+(21-1)d}=\frac{a+5d}{a+20d}$$

To make our sum equation look exactly like this term ratio, let us multiply the numerator and denominator of the target expression by $2$:

$$\frac{2(a+5d)}{2(a+20d)}=\frac{2a+10d}{2a+40d}$$

Now, compare this target format with the equation derived in Step 1:

• Compare $(p-1)d$ to $10d⟹p-1=10⟹p=11$

• Compare $(q-1)d$ to $40d⟹q-1=40⟹q=41$

Substitute these values of $p=11$ and $q=41$ directly into the right-hand side fraction $\frac{p}{q}$:

$$\frac{2a+(11-1)d}{2a+(41-1)d}=\frac{11}{41}$$

$$\frac{2a+10d}{2a+40d}=\frac{11}{41}$$

Step 3: Simplify back to find the required ratio

Factor out a $2$ from both the numerator and the denominator on the left side:

$$\frac{2(a+5d)}{2(a+20d)}=\frac{11}{41}$$

$$\frac{a+5d}{a+20d}=\frac{11}{41}$$

Since $T_6=a+5d$ and $T_{21}=a+20d$, we have:

$$\frac{T_6}{T_{21}}=\frac{11}{41}$$

Conclusion

The value of the ratio $\frac{T_6}{T_{21}}$ is $\frac{11}{41}$.

Hence, the correct option is (c) $\frac{11}{41}$.