Do you know its property? (1)

The cute property of Arithmetic Progression is :

If $\large S_{p}=S_{q}$ , & p, q are not equal

then

$\large S_{p+q}=0$

same here:

$\large S_{1729}=S_{29}$

therefore, $\large S_{1729+29}=S_{1758}=0$

$Let\quad the\quad first\quad term\quad of\quad the\quad A.P.='a'\quad and\\ it's\quad common\quad difference='d'.\\ \\ Now,{ S }_{ 1729 }={ S }_{ 29 }\\ \frac { 1729 }{ 2 } (2a+(1729-1)d)=\frac { 29 }{ 2 } (2a+(29-1)d)\\ \\ Cancelling\quad and\quad opening\quad the\quad brackets,\\ 1729(2a)+1729(1728d)\quad =\quad 29(2a)+29(28d)\\ \\ Rearranging\quad the\quad terms,\\ (1729-29)(2a)\quad +\quad (1700+29)(1700+28)d\quad =\quad 29(28d)\\ 1700(2a)+{ 1700 }^{ 2 }d+1700(29+28)d+29(28d)\quad =\quad 29(28d)\\ \\ Cancelling\quad 29(28d)\quad on\quad both\quad sides\quad \& \quad taking\quad 1700\quad common,\\ 1700\quad \{ 2a+1700d+57d\} =0\\ \\ So,\quad 2a+1757d=0\quad \quad <---------1\\ Now,\quad S_{ 1758 }=\frac { 1758 }{ 2 } (2a+(1758-1)d)\\ =\frac { 1758 }{ 2 } (2a+1757d)\\ From\quad 1,\quad 2a+1757d=0\\ \\ Then,\quad { S }_{ 1758 }=0$

CHEERS:)

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

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

That´s pretty cute to me too!!! :-)

Proof:

If $S_p = S_q$ , then

$\begin{aligned} \frac p2 (2a+(p-1)d) & = \frac q2 (2a+(q-1)d) \\ 2ap + (p^2-p)d & = 2aq + (q^2-q)d \\ 2a(p-q) + (p^2 - q^2 - p + q) d & = 0 \\ 2a(p-q) + (p-q)(p+q-1)d & = 0 \\ \implies \color{#D61F06} 2a + (p+q-1)d & = \color{#D61F06} 0 \end{aligned}$

Then $S_{p+q} = \dfrac {p+q}2 {\color{#D61F06} \big(2a + (p+q-1)d\big)} = \color{#D61F06}0$ . Since $S_{1729} = S_{29} \implies S_{1758} = S_{1729+29} = \boxed 0$ .

Actually answer should not be zero but same as first term 'a'

S29 = a + 28d and S1729 = a + 1728d

then a + 28d = a + 1728d then d = 2

so there is only one term 'a' S1758 = a + 1756 × 0 = a

S 1758 = 0, the proof is :
If Tn denotes the nth term of the given A.P, then, we may write: S29=T1 +T2 +T3 +....... +T29 and S 1729 = T1 + T2 +T3 + ......+T1729 = S 29 + (T30 +T31 +..... +T1729); Since, S 29=S1729, therefore T 30+ T31 + .... + T 1729 = 0, the 1700 terms and the middle terms are two which will be obviously equal and opposite in sign that means their sum will be equal to zero. Now, the middle terms will be T (29+1700/2 ) and T (29+(1700/2+1)) ,i.e., T 879 and T 880 . Now, suppose,T1 i.e.,first Term = a and common difference is= b, Then, T879 + T880 = 0; (a + (879-1)b ) + (a + (880-1) b) = 0, which gives: 2 a + 1757 b =0 i.e., S 1758 = 0 ( Since, S 1758 = 2a + (1758-1) b )

Sp=Sq Sp+q=0 S1729=S129 therefore S1758=0