If $7 = 5 + \frac{1}{7} (5 + \alpha) + \frac{1}{77} (5 + 2\alpha) + \frac{1}{77} (5 + 3\alpha) + \infty$, then the value of $\alpha$ is:

Step 1: Rearrange the equation

First, we isolate the infinite series by subtracting 5 from both sides of the given equation. This simplifies the problem to finding the value of $\alpha$ that makes the infinite series equal to 2.

Step 2: Split the series into two geometric progressions

We can split the given series into two separate series: one involving the constant term 5 and another involving $\alpha$. This makes it easier to identify the patterns and sum each series individually.

Step 3: Identify the first series as a geometric progression

The first series is $S_1 = \frac{1}{7} + \frac{1}{77} + \frac{1}{777} + \dots$. This is a geometric progression with first term $a = \frac{1}{7}$ and common ratio $r = \frac{1}{11}$. The sum of an infinite geometric progression is given by $S = \frac{a}{1-r}$ when $|r| < 1$.

Step 4: Calculate the sum of the first series

Using the formula for the sum of an infinite geometric progression, we substitute $a = \frac{1}{7}$ and $r = \frac{1}{11}$ to find the sum of the first series.

Step 5: Identify the second series as an arithmetico-geometric progression

The second series is $S_2 = \frac{1}{7} + \frac{2}{77} + \frac{3}{777} + \dots$. This is an arithmetico-geometric progression where the arithmetic part is $1, 2, 3, \dots$ and the geometric part is $\frac{1}{7}, \frac{1}{77}, \frac{1}{777}, \dots$. The sum of an infinite arithmetico-geometric progression $S = a + (a+d)r + (a+2d)r^2 + \dots$ is given by $S = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$. Here, $a=1$, $d=1$, and $r=\frac{1}{11}$.

Step 6: Calculate the sum of the second series

We substitute the values $a=1$, $d=1$, and $r=\frac{1}{11}$ into the formula for the sum of an infinite arithmetico-geometric progression to find $S_2$.

Step 7: Substitute sums back into the rearranged equation and solve for $\alpha$

Now we substitute the calculated sums $S_1 = \frac{11}{70}$ and $S_2 = \frac{121}{100}$ back into the equation from Step 2. Then, we solve the resulting linear equation for $\alpha$.

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