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

The Correct Option is D

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