JEE Mathematics 2025 Question with Solution

Given: The three elements of set $A$ are in A.P. and their sum is $36$. The three elements of set $B$ are also in A.P. and their sum is $36$. Also, $D = d + 3$ and

$$\frac{p+q}{p-q} = \frac{19}{5}$$

Find: The value of $p-q$.

Let the elements of set $A$ be

$$a-d,\ a,\ a+d$$

and the elements of set $B$ be

$$b-D,\ b,\ b+D$$

Since the sum of elements of set $A$ is $36$,

$$(a-d)+a+(a+d)=3a=36$$

so,

$$a=12$$

Similarly, for set $B$,

$$(b-D)+b+(b+D)=3b=36$$

so,

$$b=12$$

Therefore,

$$p=(a-d)\cdot a\cdot (a+d)=a(a^2-d^2)$$

Hence,

$$p=12(144-d^2)$$

Also,

$$q=(b-D)\cdot b\cdot (b+D)=b(b^2-D^2)$$

Hence,

$$q=12(144-D^2)$$

Using $D=d+3$,

$$q=12\left(144-(d+3)^2\right)$$

Now use the given relation

$$\frac{p+q}{p-q}=\frac{19}{5}$$

Cross-multiplying,

$$5(p+q)=19(p-q)$$ $$5p+5q=19p-19q$$ $$24q=14p$$ $$\frac{p}{q}=\frac{12}{7}$$

Substitute the expressions of $p$ and $q$:

$$\frac{144-d^2}{144-(d+3)^2}=\frac{12}{7}$$

Expanding,

$$\frac{144-d^2}{144-(d^2+6d+9)}=\frac{12}{7}$$ $$\frac{144-d^2}{135-d^2-6d}=\frac{12}{7}$$

Cross-multiplying,

$$7(144-d^2)=12(135-d^2-6d)$$ $$1008-7d^2=1620-12d^2-72d$$ $$5d^2+72d-612=0$$

Solving, we get

$$d=6$$

Since $d>0$, this value is valid. Then

$$D=d+3=9$$

Now,

$$p-q=12(144-d^2)-12(144-D^2)$$ $$p-q=12(D^2-d^2)$$

Substituting $d=6$ and $D=9$,

$$p-q=12(81-36)$$ $$p-q=12\cdot 45=540$$

Therefore, the value of $p-q$ is $540$, so the correct option is D.

Given: Both sets contain three numbers in A.P. with sum $36$, and their products are $p$ and $q$. Also, $D=d+3$ and

$$\frac{p+q}{p-q}=\frac{19}{5}$$

Find: $p-q$.

Write the A.P.s as

$$A=(12-d,\ 12,\ 12+d), \qquad B=(12-D,\ 12,\ 12+D)$$

Then

$$p=12(144-d^2), \qquad q=12(144-D^2)$$

From

$$\frac{p+q}{p-q}=\frac{19}{5}$$

we get

$$5p+5q=19p-19q$$

which gives

$$14p=24q$$

So,

$$\frac{p}{q}=\frac{12}{7}$$

Now substitute $D=d+3$:

$$\frac{12(144-d^2)}{12\left(144-(d+3)^2\right)}=\frac{12}{7}$$

Cancelling $12$,

$$\frac{144-d^2}{144-(d+3)^2}=\frac{12}{7}$$

This yields

$$7(144-d^2)=12\left(144-d^2-6d-9\right)$$ $$1008-7d^2=1620-12d^2-72d$$ $$5d^2+72d-612=0$$

Hence,

$$d=6$$

and therefore

$$D=9$$

Finally,

$$p-q=12\left[(144-d^2)-(144-D^2)\right]$$ $$p-q=12(D^2-d^2)$$ $$p-q=12(81-36)=540$$

Therefore, the correct option is D.