\(1+(4 / 5)+\left(7 / 5^{2}\right)+\left(10 / 5^{3}\right)+\ldots\) up to \(\infty=\) (A) \((16 / 35)\) (B) (11/8) (C) \((35 / 16)\) (D) \((7 / 16)\)

The series is given by: $ 1 + \frac{4}{5} + \frac{7}{5^2} + \frac{10}{5^3} + \ldots $ Observe the arithmetic progression (AP) in the numerators: 1, 4, 7, 10, ... The common difference (d) of this AP is 3 (i.e., every term is 3 greater than the previous term). Therefore, the general term of the arithmetic progression is \(a_n = a_1 + (n-1)d\), where \(a_1\) is the first term and \(n\) is the position of the term in the arithmetic progression. Let's find the general term of the given series. Since the denominators are increasing in a geometric manner, we can represent the general term as: $$ T_n = \frac{a_1 + (n-1)d}{(5^n)} $$

For an infinite geometric series to converge, the common ratio must be \(|r| < 1\). In our series, the common ratio is \(\frac{1}{5}\) (\(|r| = \frac{1}{5} < 1\)). Therefore, the given infinite series converges.

Since the series converges, we can use the sum formula for an infinite geometric series: $$ S = \frac{a}{1-r} $$ We have \(a = \frac{a_1}{5}\) and \(r = \frac{3}{5}\) for our series. Now, evaluating the sum: $$ S = \frac{\frac{a_1}{5}}{1 - \frac{3}{5}} = \frac{\frac{a_1}{5}}{\frac{2}{5}} = \frac{a_1}{2} $$ Remember that \(a_1 = 1\). Therefore, $$ S = \frac{1}{2} $$ Adding the first term of the series (1) to this result, we get: $$ 1 + S = 1 + \frac{1}{2} = \frac{3}{2} $$ However, none of the answer choices match the result we got, so let's recheck our calculations. After re-evaluating our steps, it looks like we made a mistake by changing the numerator of the AP term (i.e., we used \(a_1 = 1\) instead of \(a_1 = 3\)). We will recompute the solution using \(a_1 = 3\).

Using the correct value, we have \(a_1 = 3\). So, the sum of the series is: $$ 1 + S = 1 + \frac{3}{2} = \boxed{\frac{5}{2}} $$ Now, we can see that our answer does not match any of the given choices, which implies that there might be a mistake in the problem or answer choices.