MAH-CET 2026 — Mathematics PYQ

Let us solve this problem step-by-step using the properties of an Arithmetic Progression (AP).

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

The standard terms are given by:

$$T_r=a+(r-1)d$$

Step 1: Simplify the expression for $S_1$

$S_1$ represents the sum of all $n$ terms of the AP.

$$S_1=T_1+T_2+T_3+\cdots +T_n$$

Using the standard formula for the sum of an AP with $n$ terms:

$$S_1=\frac{n}{2}[2a+(n-1)d]$$

--- (Equation 1)

Step 2: Simplify the expression for $S_2$

$S_2$ represents the sum of the terms situated at the odd positions:

$$S_2=T_1+T_3+T_5+\cdots +T_n$$

Since $n$ is given as an odd number, the total number of terms in this sub-series is exactly $\frac{n+1}{2}$.

Let us identify the characteristics of this new sub-series:

• First term: $T_1=a$

• Common difference: The gap between consecutive terms is $T_3-T_1=2d$

• Number of terms: $n^{\prime }=\frac{n+1}{2}$

Now, apply the AP sum formula $S=\frac{n^{\prime }}{2}[2(\text{first term})+(n^{\prime }-1)(\text{common difference})]$:

$$S_2=\frac{\frac{n+1}{2}}{2}\left[2a+\left(\frac{n+1}{2}-1\right)(2d)\right]$$

$$S_2=\frac{n+1}{4}\left[2a+\left(\frac{n+1-2}{2}\right)(2d)\right]$$

$$S_2=\frac{n+1}{4}\left[2a+\left(\frac{n-1}{2}\right)(2d)\right]$$

Cancel out the $2$ inside the parentheses:

$$S_2=\frac{n+1}{4}[2a+(n-1)d]$$

--- (Equation 2)

Step 3: Find the ratio $\frac{S_1}{S_2}$

Divide Equation 1 by Equation 2:

$$\frac{S_1}{S_2}=\frac{\frac{n}{2}[2a+(n-1)d]}{\frac{n+1}{4}[2a+(n-1)d]}$$

Notice that the entire term $[2a+(n-1)d]$ is present in both the numerator and the denominator, so it cancels out completely:

$$\frac{S_1}{S_2}=\frac{\frac{n}{2}}{\frac{n+1}{4}}$$

Simplify the complex fraction:

$$\frac{S_1}{S_2}=\frac{n}{2}\times \frac{4}{n+1}$$

$$\frac{S_1}{S_2}=\frac{2n}{n+1}$$

Conclusion

The ratio $\frac{S_1}{S_2}$ evaluates to $\frac{2n}{n+1}$.

Hence, the correct option is (a) $\frac{2n}{n+1}$.