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

To solve this problem, we are given two conditions based on the infinite geometric progression (G.P.):

• The sum of the infinite G.P. is \(\sum_{n=0}^{\infty} ar^n = 57\).
• The sum of the cube terms of each term in the G.P. is \(\sum_{n=0}^{\infty} a^3 r^{3n} = 9747\).

We will use the formula for the sum of an infinite geometric series, which is:

\(S = \frac{a}{1-r}\) where \(|r| < 1\).

1. Applying the sum formula to the first series:

• \(\frac{a}{1-r} = 57\).
• Let's denote this as Equation (1).

2. Applying the sum formula to the second series:

Note that the common ratio is \(r^3\) and the first term is \(a^3\):

• \(\frac{a^3}{1 - r^3} = 9747\).
• Let's denote this as Equation (2).

Now, let's solve these two equations:

From Equation (1):

• \(a = 57(1 - r)\).

Substitute \(a\) in Equation (2):

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

Simplifying gives:

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

Since \((1 - r)\) can be cancelled, we have:\)

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

Simplifying, we have:

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

Using \(57^3 = 185193\), we simplify:\)

\((1 - r)^2(1 + r + r^2) = \frac{9747}{185193}\)

Simplifying gives us:

\(1 - r^2 = \frac{1}{19}\)\), hence \(r = \frac{2}{3}\)\), \(1 - r = \frac{1}{3}\)\).

From \(a = 57(1 - r)\):

\(a = 57 \cdot \frac{1}{3} = 19\).

Finding \(a + 18r\):

\(a + 18r = 19 + 18 \cdot \frac{2}{3} = 19 + 12 = 31\).

Hence, the value of \(a + 18r\) is 31, making the correct answer 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