MAH-CET 2026 — Mathematics PYQ

Step 1: Identify the first and last terms of the series

We need to find the natural numbers between $250$ and $1000$ that leave no remainder when divided by $3$.

• Finding the first term ($a$): Divide $250$ by $3⟹\frac{250}{3}=83.33$.

  The next complete multiple of $3$ is $84\times 3=252$.

  Therefore, the first term $a=252$.

• Finding the last term ($l$): The question specifies numbers between $250$ and $1000$, so we test numbers below $1000$.

  Divide $999$ by $3⟹\frac{999}{3}=333$ (exactly divisible).

  Therefore, the last term $l=999$.

The sequence forms an Arithmetic Progression (A.P.):

$$252,255,258,\dots ,999$$

Where the common difference ($d$) is $3$.

Step 2: Find the total number of terms ($n$)

The formula for the $n$-th term ($a_n$) of an Arithmetic Progression is:

$$a_n=a+(n-1)d$$

Substitute the values $a_n=999$, $a=252$, and $d=3$:

$$999=252+(n-1)3$$

Subtract $252$ from both sides:

$$999-252=(n-1)3$$

$$747=(n-1)3$$

Divide both sides by $3$:

$$n-1=\frac{747}{3}$$

$$n-1=249$$

$$n=249+1=250$$

So, there are $250$ terms in this series.

Step 3: Calculate the sum of the series ($S_n$)

The formula for the sum of an Arithmetic Progression when the first and last terms are known is:

$$S_n=\frac{n}{2}(a+l)$$

Substitute $n=250$, $a=252$, and $l=999$:

$$S_{250}=\frac{250}{2}(252+999)$$

$$S_{250}=125\times 1251$$

Let us perform the multiplication:

$$S_{250}=156375$$

Correct Answer:

(a) 156375