Which term?

I totally Like your Solution @Ashish Siva , but we can Reduce the calculation .

We have an A.P of Common Difference $(d) = \dfrac{17}{8}-\dfrac58=\dfrac{12}8$ and First Term $(a) = \dfrac{5}8$ . Let the Required number of terms be $n$

$S_n=\dfrac{295}{4}=\dfrac{n}{2} \left(\dfrac54+(n-1)\dfrac64\right) \\ 590=n(5+6n-6) \\ n(6n-1)=590 \\ n(6n-1)=10(6\times10-1) \\ \boxed{n = 10 }$

Since $d>0$ , the quadratic will only give us only $1$ +ve Solution

Here we see that the first term of the AP(a) is $\dfrac{5}{8}$ and the common difference(d) = $\dfrac{17 - 5}{8} = \dfrac{12}{8} = \dfrac{3}{2}$ . Let the number of terms required be $n$ .

$S_n = \dfrac{n}{2} × \left(2a + (n - 1)d\right)\\ \dfrac{295}{4} = \dfrac{n}{2} × \left(2×\dfrac{5}{8} + (n - 1)×\dfrac{3}{2}\right)\\ \dfrac{295}{4} = \dfrac{n}{2} × \left(\dfrac{5}{4} + \dfrac{3n - 3}{2}\right)\\ \dfrac{295}{4} = \dfrac{n}{2} × \dfrac{5 + 6n - 6}{4}\\ \dfrac{295}{4} = \dfrac{6n^2 - n}{8}\\ \dfrac{295 × 8}{4} = 6n^2 - n\\ 590 = 6n^2 - n\\ 6n^2 - n - 590 = 0$

Applying quadratic formula:-
$n = \dfrac{ -(-1) \pm \sqrt{{(-1)}^2 -4×6×-590}}{2×6}\\ \\ n = \dfrac{1 \pm \sqrt{1 + 14160}}{12}\\ \\ n = \dfrac{1 \pm \sqrt{14161}}{12}\\ \\ n = \dfrac{1 \pm 119}{12}\\ \\ n = \dfrac{1 + 119}{12} \left(\text{or}\right) \dfrac{1 - 119}{12}\\ \\ n = \dfrac{120}{12} \left(\text{or}\right) = \dfrac{-118}{12}\\ \\ n = 10 \left(\text{or}\right) \dfrac{-59}{6}$

But, number of terms can never be negative. So, $n \neq \dfrac{-59}{6}$ . So, $n = \color{#69047E}{\boxed{10}}$ .

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Alternatively:-
Observe that the numerators of the terms are in AP with first term(a) = $5$ and common difference $12$ . Why? Because the actual common difference $\dfrac{3}{2}$ can be written as $\dfrac{12}{8}$ . So denominator remains the same but numerator increases by $12$ .

The sum given whihc is $\dfrac{295}{4}$ can be written as $\dfrac{590}{8}$ . So, now we just have to find how much terms of the AP $5, 17, 29, 41, \cdots$ give the sum as $590$ . Let $m$ terms be required.
$S_m = \dfrac{m}{2} × \left(2a + (m - 1)d\right)\\ 590 = \dfrac{m}{2} × \left(2×5 + (m - 1)×12\right)\\ 590 = \dfrac{m}{2} × 2\left(5 + 6(m - 1)\right)\\ 590 = m\left(5 + 6m - 6\right)\\ 590 = 6m^2 - m\\ 6m^2 - m - 590 = 0\\ 6m^2 - 60m + 59m - 590 = 0\\ 6m\left(m - 10\right) + 59\left(m - 10\right) = 0\\ \left(m - 10\right)\left(6m + 59\right) = 0\\ m - 10 = 0\left(\text{or}\right) 6m + 59 = 0\\ n = 10 \left(\text{or}\right) \dfrac{-59}{6}$

But, number of terms can never be negative. So, $n \neq \dfrac{-59}{6}$ . So, $n = \color{#69047E}{\boxed{10}}$ .