JEE Mathematics 2025 Question with Solution

Given:

$$7 = 5 + \frac{1}{7}(5 + \alpha) + \frac{1}{7^2}(5 + 2\alpha) + \frac{1}{7^3}(5 + 3\alpha) + \cdots$$

Find: The value of $\alpha$.

The solution states that this is an infinite series with common ratio $\frac{1}{7}$ and concludes that the correct option is A with final answer $\alpha = \frac{6}{7}$.

A more detailed working shown on the page rewrites the series and obtains $\alpha = 6$, then explicitly notes a discrepancy and finally states that the answer for $\alpha$ that satisfies the equation is

$$\frac{6}{7}$$

Thus, based on the solution, the correct option is A.

Therefore, the value of $\alpha$ is $\frac{6}{7}$.

Given:

$$7 = 5 + \frac{1}{7}(5 + \alpha) + \frac{1}{7^2}(5 + 2\alpha) + \frac{1}{7^3}(5 + 3\alpha) + \cdots$$

Find: The value of $\alpha$.

The extracted detailed solution rewrites the expression as

$$7 = 5 + \sum_{n=1}^{\infty} \frac{5 + n\alpha}{7^n}$$

and separates it into two series:

$$\sum_{n=1}^{\infty} \frac{5 + n\alpha}{7^n} = 5\sum_{n=1}^{\infty} \frac{1}{7^n} + \alpha \sum_{n=1}^{\infty} \frac{n}{7^n}$$

Using

$$\sum_{n=1}^{\infty} x^n = \frac{x}{1-x}$$

and

$$\sum_{n=1}^{\infty} n x^n = \frac{x}{(1-x)^2}$$

with $x = \frac{1}{7}$, it computes

$$\sum_{n=1}^{\infty} \frac{1}{7^n} = \frac{1}{6}$$

and

$$\sum_{n=1}^{\infty} \frac{n}{7^n} = \frac{7}{36}$$

Substituting back gives

$$7 = 5 + 5 \times \frac{1}{6} + \alpha \times \frac{7}{36}$$

so

$$7 = 5 + \frac{5}{6} + \frac{7\alpha}{36}$$

Then

$$7 - 5 - \frac{5}{6} = \frac{7\alpha}{36}$$ $$\frac{7}{6} = \frac{7\alpha}{36}$$

which leads to

$$\alpha = 6$$

The same extracted solution then says the answer in the problem is $\frac{6}{7}$, mentions a possible misread or inconsistency, and finally boxes

$$\frac{6}{7}$$

Hence, the solution is internally inconsistent: its working yields $6$, but its stated final answer and option marker are $\frac{6}{7}$. Following the instruction that the solution is the primary source for the answer, we retain the stated final answer from the solution's.

Therefore, the correct option is A, corresponding to $\frac{6}{7}$.