NIMCET 2025 — Mathematics PYQ

Let the first term of the Arithmetic Progression (AP) be $a$ and its common difference be $d$.

The $n$-th term of the AP is given by the formula:

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

Step 1: Simplify the first given condition

We are given:

$$\sum _{l=0}^{18}{t}_{3l+1}=1197$$

Expanding the summation by putting $l=0,1,2,\dots ,18$:

$$t_1+t_4+t_7+\dots +t_{55}=1197$$

This is a new sequence of $19$ terms ($l$ goes from $0$ to $18$, so there are $18-0+1=19$ terms).

• The first term is $t_1=a$.

• The last term is $t_{55}=a+54d$.

Using the sum formula for an AP, $S_n=\frac{n}{2}[\text{first term}+\text{last term}]$:

$$\frac{19}{2}(t_1+t_{55})=1197$$

Divide both sides by $19$ (note that $1197÷19=63$):

$$\frac{1}{2}(a+a+54d)=63$$

$$\frac{1}{2}(2a+54d)=63$$

$$a+27d=63\text{— (Equation 1)}$$

Step 2: Simplify the second given condition

We are given:

$$t_7+3t_{22}=174$$

Expressing these terms in terms of $a$ and $d$:

$$(a+6d)+3(a+21d)=174$$

$$a+6d+3a+63d=174$$

$$4a+69d=174\text{— (Equation 2)}$$

Step 3: Solve Equation 1 and Equation 2

From Equation 1, we can express $a$ as:

$$a=63-27d$$

Substitute this value of $a$ into Equation 2:

$$4(63-27d)+69d=174$$

$$252-108d+69d=174$$

$$252-39d=174$$

$$252-174=39d$$

$$78=39d$$

$$d=2$$

Now, substitute $d=2$ back into Equation 1 to find $a$:

$$a=63-27(2)$$

$$a=63-54$$

$$a=9$$

So, the first term $a=9$ and the common difference $d=2$.

Step 4: Calculate the value of ${\sum }_{l=1}^9{t}_l^2$

We need to evaluate the sum of the squares of the first $9$ terms:

$$\sum _{l=1}^9{t}_l^2=t_1^2+t_2^2+t_3^2+\dots +t_9^2$$

Let's write down the first $9$ terms using $a=9$ and $d=2$:

• $t_1=9$

• $t_2=11$

• $t_3=13$

• $\dots$

• $t_9=9+8(2)=25$

Thus, we need to find the sum of squares of consecutive odd numbers from $9$ to $25$:

$$\sum _{l=1}^9{t}_l^2=9^2+11^2+13^2+15^2+17^2+19^2+21^2+23^2+25^2$$

Calculating each square:

• $9^2=81$

• $11^2=121$

• $121+81=202$

• $13^2=169$

• $202+169=371$

• $15^2=225$

• $371+225=596$

• $17^2=289$

• $596+289=885$

• $19^2=361$

• $885+361=1246$

• $21^2=441$

• $1246+441=1687$

• $23^2=529$

• $1687+529=2216$

• $25^2=625$

• $2216+625=2841$

So, the total sum is:

$$\sum _{l=1}^9{t}_l^2=2841$$

Step 5: Find the value of $b$

We are given that:

$$\sum _{l=1}^9{t}_l^2=947b$$

Substitute the calculated sum into the equation:

$$2841=947b$$

$$b=\frac{2841}{947}$$

$$b=3$$

Correct Answer:

The value of $b$ is 3 (Option C).