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:

Question

If $ A = \begin{bmatrix} 1 & 0 \\ 1/2 & 1 \end{bmatrix} $, then $ A^{50} $ is:

Answer 1

To resolve this issue, two conditions derived from an infinite geometric progression (G.P.) are provided:

The sum of the infinite G.P. is expressed as $\sum_{n=0}^{\infty} ar^n = 57$.
The sum of the cubes of each term in the G.P. is given by $\sum_{n=0}^{\infty} a^3 r^{3n} = 9747$.

The formula for the sum of an infinite geometric series, applicable when $|r| < 1$, is:

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

1. Applying the sum formula to the initial series yields:

2. Applying the sum formula to the second series, with a common ratio of $r^3$ and a first term of $a^3$:

Solving these two equations:

From Equation (1):

Substituting $a$ into Equation (2):

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

Upon simplification:

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

After canceling $(1 - r)$:

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

Further simplification results in:

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

Utilizing $57^3 = 185193$, the equation simplifies to:

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

This simplifies to $1 - r^2 = \frac{1}{19}$, yielding $r = \frac{2}{3}$ and $1 - r = \frac{1}{3}$.

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

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

Calculating $a + 18r$:

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

The value of $a + 18r$ is 31, confirming 31 as the correct answer.

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: