Let $$S_{n}$$ be sum of the first n terms of an A.P. If $$S_{5} = S_{9}$$, what is the ratio of $$a_3:a_5$$

We are given that $$S_{5} = S_{9}$$

We know the sum of n terms of a given AP is

$$S_n=\dfrac{n}{2}\left[2a+\left(n-1\right)d\right]$$

Where a is the first term, d is the common difference, and n is the number of terms.

Equating $$S_{5} = S_{9}$$

$$\dfrac{5}{2}\left[2a+4d\right]=\dfrac{9}{2}\left[2a+8d\right]$$

$$10a+20d=18a+72d$$

$$-8a=52d$$

$$a=-\dfrac{52d}{8}=-\dfrac{13d}{2}$$

Now we need to find the ratio of $$a_3:a_5$$

$$\dfrac{a_3}{a_5}=\dfrac{a+2d}{a+4d}=\dfrac{-\dfrac{13d}{2}+2d}{-\dfrac{13d}{2}+4d}$$

$$\dfrac{a_3}{a_5}=\dfrac{-\dfrac{9d}{2}}{-\dfrac{5d}{2}}=\dfrac{9}{5}$$

9:5