My take on the question -
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First thing, we've to notice few things-
- the elements of $T$ are in Arithmetic Progression or AP.
- the sum of the first term & the last term is $470$
- Similarly, the sum of the $2^{nd}$ term & the $2^{nd}$ last term is $470$ and the pattern continues ......
Now, as we know,
the formula to find $n^{th}$ term of an AP series
$\qquad a+(n-1)d = t_n$
$\qquad where, a = \text{first term,}$
$\qquad \text{n=no. of terms,}$
$\qquad \text{d=common difference,}$
$\qquad t_n = n^{th} term$
∴ from this formula, we can determine the no. of elements in the set $T$.
in set $T$ -
$\qquad \text{first term (a) = 3}$
$\qquad \text{common difference between elements = (d) = 8} $
$\qquad n^{th} \text{term = 467}$
$\qquad \text{no. of terms(n) = ?}$
∴ $467 = 3+(n-1)8$
$\Rightarrow 467 = 3+8n-8$
$\Rightarrow 467 = 8n-5$
$\Rightarrow 467+5 = 8n$
$\Rightarrow n = \dfrac{472}{8} = 59$
∴ In set $T$, there are a total of $59$ elements.
∴ In set $T$, there will be $\left (\dfrac{59}{2} = 29 \right ) pairs$ (if we add the elements in each pair, then we'll get $470$) , & only $1$ element left i.e. $\left (\dfrac{3+467}{2} = \dfrac{470}{2} = 235\right) $.
From set $T$, we've to form a subset $S$ where the sum of no two elements will be $470$.
∴ We can take $1$ element from $29$ pairs i.e. $29$ elements & place them to set $S$.
And we can also place the only element $( i.e. 235)$ which has no pair, to $S$.
∴ $\color{violet}{\text{Set S can have maximum}}$ $\color{maroon}{30}$ $\color{violet}{\text{elements in it.}}$
