Let $S_{n}$ be sum of the first ' n ' terms of an A.P. $\left\{ a_{n} \right\}$. If $S_{6} = S_{9}$,

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Let $S_{n}$ be sum of the first ' n ' terms of an A.P. $\left\{ a_{n} \right\}$. If $S_{6} = S_{9}$, then the ratio $a_{3}:a_{5}$ is equal to

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Sol. [Key Concept: Use $S_{n} = \frac{\ }{n}2(2a + (n - 1)d)$ and $a_{k} = a + (k - 1)d$; the given $S$-ratio fixes the term ratio.]$$\frac{6}{2}(2a + 5	ext{ }d) = \frac{9}{2}(2a + 8	ext{ }d)$$$$\begin{matrix} \Rightarrow 4a + 10d & \ = 6a + 24d \Rightarrow a + 7d = 0 \ \frac{a_{3}}{a_{5}} & \ = \frac{a + 2d}{a + 4d} = \frac{- 5d}{- 3d} = \frac{5}{3} \end{matrix}$$

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