Do you know its property - 2

${ S }_{ p }={ S }_{ q }\quad \Rightarrow \cfrac { p }{ 2 } \left\{ 2a+(p-1)d \right\} =\cfrac { q }{ 2 } \left\{ 2a+(q-1)d \right\} \\ \\ \Rightarrow (p-q)a+\cfrac { 1 }{ 2 } ({ p }^{ 2 }-{ q }^{ 2 }-p+q)d=0\\ \\ \Rightarrow a+\cfrac { p+q-1 }{ 2 } d=0\quad \\ \\ \Rightarrow a+(\cfrac { p+q+1 }{ 2 } -1)d=0\\ \\ \Rightarrow { T }_{ \cfrac { p+q+1 }{ 2 } }=0\quad (Only\quad if\quad \cfrac { p+q+1 }{ 2 } \in { I }^{ + })$

$S_{1729} = S_{28}\quad \Rightarrow \dfrac {1729(2a+1728d)}{2} = \dfrac {28(2a+27d)}{2}$

$\Rightarrow 1729(2a+1728d) = 28(2a+27d)$

$\Rightarrow 2a(1729-28) = [28(27)-1729(1728)]d$

$\Rightarrow 3402a =-2986956d \quad \Rightarrow a = -878d$

Now, the $879^{th}$ term $a_{879} = a + 878d = -878d+878d = \boxed{0}$

Geometrical Approach

Again if in an AP if $S_p=S_q$ for $q>p$ then $\Large a_{\frac{p+q+1}{2}}=0$ (This is valid only when $\frac{p+q+1}{2}$ is an integer).

You can simply visualize this using the graph of the AP. The AP line whether it has positive slope or the negative, will make two congruent triangle one above & the other below between $\large n_1=p+1$ & $\large n_2=q$ around $\large n_0=\frac{p+q+1}{2}$ , therby meeting the horizontal axis at $n_0$ in both the cases.

Hence $\huge a_{\frac{1729+28+1}{2}}=a_{879}=0$

I am writing for the long, basic expandded method for begginers. Let the AP be a, a+d, a+2d etc. so, 28/2[2a+(27)d] = 1729/2[2a+(1728)d] After expanding, we get: 56a = 756d = 3458a + 2987712d . (Trust me, if you have no other plan , do this, because 1728*1729 would be a very very tedious process) After simplifications, we get: 3402a + 2896956d = 0. (Dividing throughout by 3402, we get) a + 878d = 0 So, 879th term is 0.

An arithmetic progression is either increasing by a constant, decreasing by a constant, or always a constant. If the progression is increasing, then the sum should be increasing, thus the sum of the first 1729 terms can't be equal to the sum of the first 28. Likewise, if the progression is decreasing, then the sum should be decreasing, thus the sum of the first 1729 terms can't be equal to the sum of the first 28 too! So the progression is constant. But if the sum of the 1729 terms is equal to the sum of the 28 terms, there are three cases: (1) If the initial value is positive, then the sum should be increasing. (2) If the initial value is negative, then the sum should be decreasing. (3) If the initial value is 0, then the sum should be constant. Thus, the sequence is 0, 0, 0, 0, 0, ... or the nth term is always 0. The 879th term is 0. Q.E.D

For the given equation to be true the sum of terms after 28th term has to be zero. i.e 29th to 1729th term....total 1701 terms. If the sum of an odd no. of terms of an AP are zero, then the middle term has to be zero. therefore, starting from 29th term, the middle term is 851th term. Starting from 1st term, it is 851+28 = 879th term = 0

What Sandeep Bhardwaj told us in his solution, actually, for any AP is that

$\large S_{n+k}=\left(\frac{n+k}{n-k}\right)\left({ S}_{n}-{S}_{k}\right) \quad \quad n>k$

Let´s take a look another fact,

$\large S_{2m-1}=(2m-1)a_m$

For $n=1729$ , $k=28$ and $m=879$

$\Large a_{879}=0$

take all terms========0