JEE Mathematics 2024 Question with Solution

Given:

• $\sum_{n=0}^{\infty} ar^n = 57$
• $\sum_{n=0}^{\infty} a^3r^{3n} = 9747$

Find: $a + 18r$

For an infinite G.P.,

$$S = \frac{a}{1-r}$$

with $|r| < 1$.

From the first series,

$$\frac{a}{1-r} = 57$$

so

$$a = 57(1-r)$$

For the second series, the first term is $a^3$ and the common ratio is $r^3$. Hence,

$$\frac{a^3}{1-r^3} = 9747$$

Substitute $a = 57(1-r)$:

$$\frac{(57(1-r))^3}{1-r^3} = 9747$$

Using $1-r^3 = (1-r)(1+r+r^2)$,

$$\frac{57^3(1-r)^3}{(1-r)(1+r+r^2)} = 9747$$

Thus,

$$\frac{57^3(1-r)^2}{1+r+r^2} = 9747$$

Now $57^3 = 185193$ and $9747 = \frac{185193}{19}$, so

$$\frac{(1-r)^2}{1+r+r^2} = \frac{1}{19}$$

Therefore,

$$19(1-r)^2 = 1+r+r^2$$

Solving gives $r = \frac{2}{3}$.

Then

$$a = 57\left(1-\frac{2}{3}\right) = 19$$

So,

$$a + 18r = 19 + 18\cdot\frac{2}{3} = 31$$

Therefore, the correct option is D.

Given:

• $\frac{a}{1-r} = 57$
• $\frac{a^3}{1-r^3} = 9747$

Find: $a + 18r$

From

$$a = 57(1-r)$$

substitute into the second equation:

$$\frac{57^3(1-r)^3}{1-r^3} = 9747$$

Using $1-r^3 = (1-r)(1+r+r^2)$,

$$\frac{57^3(1-r)^2}{1+r+r^2} = 9747$$

Since $\frac{9747}{57^3} = \frac{1}{19}$,

$$\frac{(1-r)^2}{1+r+r^2} = \frac{1}{19}$$

So,

$$19(1-2r+r^2) = 1+r+r^2$$ $$19 - 38r + 19r^2 = 1 + r + r^2$$ $$18r^2 - 39r + 18 = 0$$ $$6r^2 - 13r + 6 = 0$$ $$(3r-2)(2r-3) = 0$$

Hence $r = \frac{2}{3}$ or $r = \frac{3}{2}$. Since the G.P. is infinite, $|r| < 1$, so

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

Then

$$a = 57\left(1-\frac{2}{3}\right) = 19$$

Finally,

$$a + 18r = 19 + 12 = 31$$

Thus, the required value is $31$, so the correct option is D.