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For a certain arithmetic progression, the sum of the first 1729 terms is equal to the sum of the first 29 terms.  Find, with proof, the sum of the first 1758 terms.  Show all your steps!

Dec 3, 2019

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For a certain arithmetic progression, the sum of the first 1729 terms is equal to the sum of the first 29 terms.

Find, with proof, the sum of the first 1758 terms.

Show all your steps!

Formula arithmetic progeression: $$a_n = a_1 + (n-1)d$$

$$\begin{array}{|rcll|} \hline s_{1729} &=& \left(\dfrac{a_1+a_{1729}}{2}\right)*1729 \quad & | \quad a_{1729} = a_1 + 1728d \\ s_{1729} &=& \left(\dfrac{a_1+a_1 + 1728d}{2}\right)*1729 \\ s_{1729} &=& \left(\dfrac{2a_1+ 1728d}{2}\right)*1729 \quad & | \quad \text{the sum of the first 1729 terms} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline s_{29} &=& \left(\dfrac{a_1+a_{29}}{2}\right)*29 \quad & | \quad a_{29} = a_1 + 28d \\ s_{29} &=& \left(\dfrac{a_1+a_1 + 28d}{2}\right)*29 \\ s_{29} &=& \left(\dfrac{2a_1+ 28d}{2}\right)*29 \quad & | \quad \text{the sum of the first 29 terms} \\ \hline \end{array}$$

$$\mathbf{s_{1729}=s_{29}}$$

$$\begin{array}{|rcll|} \hline \mathbf{s_{1729} } &=& \mathbf{s_{29}} \\\\ \left(\dfrac{2a_1+ 1728d}{2}\right)*1729 &=& \left(\dfrac{2a_1+ 28d}{2}\right)*29 \\ \left( 2a_1+ 1728d \right)*1729 &=& \left( 2a_1+ 28d \right)*29 \\ 2*1729a_1+1728*1729d &=& 2*29a_1+28*29d \\ 2a_1(1729-29) &=& (28*29-1728*1729)d \\ 2a_1*1700 &=& -2986900d \\ 2a_1*17 &=& -29869d \\ \mathbf{ 2a_1 } &=& \mathbf{-1757d} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline s_{1758} &=& \left(\dfrac{a_1+a_{1758}}{2}\right)*1758 \quad & | \quad a_{1758} = a_1 + 1757 \\ s_{1758} &=& \left(\dfrac{a_1+a_1 + 1757d}{2}\right)*1758 \\ s_{1758} &=& \left(\dfrac{2a_1+ 1757d}{2}\right)*1758 \quad & | \quad \text{the sum of the first 1758 terms} \\ && \boxed{\mathbf{ 2a_1 = -1757d}} \\ s_{1758} &=& \left(\dfrac{-1757d+ 1757d}{2}\right)*1758 \\ s_{1758} &=& \left(\dfrac{0}{2}\right)*1758 \\ \mathbf{s_{1758}} &=& \mathbf{0} \\ \hline \end{array}$$

The sum of the first 1758 terms is 0.

Dec 3, 2019