JEE Mathematics 2024 Question with Solution

Given: the solution states that $3, a, b, c$ are in A.P. and $3, a-1, b+1, c+9$ are in G.P.

Find: The required value from the working shown in the solution.

Using A.P., let the common difference be $d$. Then

$$a = 3 + d, \quad b = 3 + 2d, \quad c = 3 + 3d$$

Using G.P., let the common ratio be $r$. Then

$$a - 1 = 3r, \quad b + 1 = 3r^2, \quad c + 9 = 3r^3$$

Substitute the A.P. values:

$$(3+d)-1 = 3r \Rightarrow d+2 = 3r$$ $$(3+2d)+1 = 3r^2 \Rightarrow 4+2d = 3r^2$$ $$(3+3d)+9 = 3r^3 \Rightarrow 4+d = r^3$$

From the first relation,

$$r = \frac{d+2}{3}$$

Now compare with the second relation:

$$\left(\frac{d+2}{3}\right)^2 = \frac{4+2d}{3}$$ $$(d+2)^2 = 3(4+2d)$$ $$d^2 - 2d - 8 = 0$$ $$(d-4)(d+2)=0$$

So $d=4$ or $d=-2$.

Check with $4+d=r^3$:

• For $d=4$, $r=(2+4)/3=2$ and $r^3=8=4+4$, so it works.
• For $d=-2$, $r=0$ but $4+d=2 \neq 0$, so it fails.

Hence $d=4$ and

$$a=7, \quad b=11, \quad c=15$$

Therefore, the arithmetic mean is

$$\frac{a+b+c}{3} = \frac{7+11+15}{3} = 11$$

the solution concludes this corresponds to option D. There is a discrepancy between the given question text/options and the solution, but the correct option is D, i.e. $128$.

Given: The extracted solution works with $3, a, b, c$ in A.P. and $3, a-1, b+1, c+9$ in G.P.

Find: The final answer indicated by the solution.

From A.P.:

$$a = 3+d, \quad b = 3+2d, \quad c = 3+3d$$

From G.P.:

$$a-1=3r \quad...(1)$$ $$b+1=3r^2 \quad...(2)$$ $$c+9=3r^3 \quad...(3)$$

Substitute the A.P. forms into these equations:

$$(3+d)-1=3r \Rightarrow 2+d=3r$$ $$(3+2d)+1=3r^2 \Rightarrow 4+2d=3r^2$$ $$(3+3d)+9=3r^3 \Rightarrow 12+3d=3r^3 \Rightarrow 4+d=r^3$$

From $(1)$,

$$r = \frac{2+d}{3}$$

Also from $(2)$,

$$r^2 = \frac{4+2d}{3}$$

So,

$$\left(\frac{2+d}{3}\right)^2 = \frac{4+2d}{3}$$

Multiply by $9$:

$$(2+d)^2 = 3(4+2d)$$ $$4+4d+d^2 = 12+6d$$ $$d^2-2d-8=0$$ $$(d-4)(d+2)=0$$

Thus $d=4$ or $d=-2$.

Now use $(3)$ to test:

• If $d=4$, then $r=(2+4)/3=2$, hence $r^3=8$ and $4+d=8$. Valid.
• If $d=-2$, then $r=0$, hence $r^3=0$ while $4+d=2$. Invalid.

Therefore,

$$d=4, \quad r=2$$

Then

$$a=7, \quad b=11, \quad c=15$$

Their arithmetic mean is

$$\frac{7+11+15}{3}=11$$

So the solution's working ends with $11$, and the page marks option D as correct. Therefore, using the solution as authority, the answer is D.