Question

Let $a, ar, ar^2, \dots$ be an infinite G.P. If $\sum_{n=0}^\infty ar^n = 57 \quad \text{and} \quad \sum_{n=0}^\infty a^3 r^{3n} = 9747,$ then $a + 18r$ is equal to:

• A. 27
• B. 46
• C. 38
• D. 31

Answer 1

Answer: (D)

31

Answer 2

We are given the following information about the infinite geometric series:

1. $\sum_{n=0}^{\infty} ar^n = 57$
2. $\sum_{n=0}^{\infty} a3^nr^n = 9747$

We need to find the value of $a + 18r$.

Step 1: Use the formula for the sum of an infinite geometric series

The sum of an infinite geometric series $\sum_{n=0}^{\infty} ar^n$ is given by the formula:

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

From the first given equation:

$\sum_{n=0}^{\infty} ar^n = 57$

Substitute this into the formula:

$\frac{a}{1 - r} = 57 \implies a = 57(1 - r) \quad \text{(Equation I)}$

Step 2: Use the second geometric series sum

Next, we are given the second series:

$\sum_{n=0}^{\infty} a3^nr^n = 9747$

This is a geometric series with the first term $a3^r$ and common ratio 3. The sum of the infinite series is:

$S = \frac{a3^r}{1 - 3} = \frac{a3^r}{-2}$

Substitute this into the given equation:

$\frac{a3^r}{-2} = 9747 \implies a3^r = -2 \times 9747 = -19494 \quad \text{(Equation II)}$

Step 3: Solve the system of equations

Now, we have two equations:

1. $a = 57(1 - r)$
2. $a3^r = -19494$

Substitute Equation I into Equation II:

$57(1 - r)3^r = -19494$

Simplify:

$(1 - r)3^r = \frac{-19494}{57} = -342$

Now, cube both sides of Equation I to eliminate $r$:

$(1 - r)^3 = \frac{57^3}{9717} = 19$

Thus:

$(1 - r)^3 = 19 \implies 1 - r = \frac{2}{3}$

So:

$r = 1 - \frac{2}{3} = \frac{1}{3}$

Step 4: Calculate $a$ and $a + 18r$

Now substitute $r = \frac{2}{3}$ back into Equation I:

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

Now, calculate $a + 18r$:

$a + 18r = 19 + 18 \times \frac{2}{3} = 19 + 12 = 31$

Thus, the correct answer is:

31