The sum of the first 24 terms of an AP is $$\frac{12504}{25}$$ and the sum of the next 24 terms is $$\frac{17112}{25}$$. What is the $$3^{rd}$$ terms?

The formulae for the sum of first n terms in AP is $$\frac{n}{2}\cdot\left(2\cdot a+\left(n-1\right)\cdot d\right)$$.

$$\frac{12504}{25}=\frac{24}{2}\cdot\left(2\cdot a+23\cdot d\right)$$

$$\frac{1042}{25}=\left(2\cdot a+23\cdot d\right)$$

Lets generate other equation, given the sum of next 24 terms is $$\frac{17112}{25}$$.

So the first 48 terms is $$\frac{29616}{25}=\frac{48}{2}\cdot\left(2a+\left(47\cdot d\right)\right)$$

Solving both we get a=429/25 d=8/25